math248ex1 (combi)

Mathematics 248 Topics in Combinatorial Mathematics 1st Semester AY 2006-2007

1. Find the number of ways to choose a pair {a, b} of distinct numbers from the set {1, 2, . . . , 50} suchthat

(i) |a− b| = 5

(ii) |a− b| ≤ 5

2. There are 12 students in a party. Five of them are girls. In how many ways can these 12 studentsbe arranged in a row if

(i) there are no restrictions?JS The number of ways to arrange the 12 students is 12! = P12

12.(ii) the 5 girls must be together (forming a block)?

JS The number of ways to arrange five girls within a block is 5!, while the number of ways toarrange 8 objects — 7 boys and 1 block of girls — is 8!, thus by M.P., the number of ways thatthe students can be arranged is 5! · 8!.

(iii) no 2 girls are adjacent?JS Given that H12

5 =(12−5+1

5

)is the number of ways that five places can be chosen amongst

twelve, so that no two of the places are consecutive, which will be the positions for the girls,5! ways to arrange the five girls amongst those positions and 7! ways to arrange the sevenboys amongst the remaining positions, by M.P., the number of ways that the students can bearranged is 5! · 7! ·

(85

).

(iv) between two particular boys A and B, there are no boys but exactly 3 girls?JS There are 2! ways to arrange A and B,

(53

)ways to choose the three girls amongst the

five that would be between A and B, 3! to arrange these three chosen girls, and 8! ways toarrange the block of five students (A, B and the three girls between them) with the remainingseven students. By M.P., the number of ways that the students can be arranged is 2!·

(53

)·3!·8!.

3. m boys and n girls are to be arranged in a row, where m,n ∈ N. Find the number of ways this canbe done in each of the following cases:

(i) There are no restrictions;JS The number of ways to arrange m + n people in a row is (m + n)! = Pm+n

m+n.(ii) No boys are adjacent (m ≤ n + 1);

JS There are Hm+nm =

(m+n−m+1

m

)ways of choosing m places among m+n such that no two

of these places are consecutive, which will be the positions of the boys, m! ways to arrangethe boys among the chosen positions and n! ways to arrange the girls among the remainingpositions, by M.P., the number of ways to arrange the students is m! · n! ·

(n+1m

).

(iii) The n girls form a single block;JS There are n! ways to arrange the girls within their block and (m + 1)! ways to arrangem + 1 objects — the m boys and the block of girls — by M.P., the number of ways to arrangethe students is n! · (m + 1)!.

(iv) A particular boy and a particular girl must be adjacent.JS There are 2! ways to arrange the boy and the girl within their block, and (m+n− 1)! waysto arrange m + n − 2 + 1 objects — the boy-and-girl block and the m + n − 2 other students— by M.P., the number of ways to arrange the students is 2! · (m + n− 1)!.

4. How many 5-letter words can be formed using A, B, C, D, E, F, G, H, I, J,

(i) if the letters in each word must be distinct?JS The number of arrangements of five distinct letters from among the ten letters is P10

5 =10!5! = 10 · 9 · 8 · 7 · 6.

(ii) if, in addition, A, B, C, D, E, F can only occur as the first, third of fifth letters while the rest asthe second or fourth letters?JS For the odd-numbered letters, the choices for the letters are 6, 5 and 4, respectively; forthe even-numbered letters, the choices for the letters are 4 and 3, repectively. By M.P., thenumber of five-letter words that can be constructed is 6 · 4 · 5 · 3 · 4.

Exercises 1 of 32


5. Find the number of ways of arranging the 26 letters in the English alphabet in a row such thatthere are exactly 5 letters between x and y.

6. Find the number of odd integers between 3000 and 8000 in which no digit is repeated.

JS There are two types of numbers that satisfy the conditions:

Type 1: Numbers with odd thousands digits: the thousands digit must be 3, 5 or 7. This will leaveonly 4 possible digits for the ones digit, and 8 · 7 choices for the middle two digits. There are3 · 8 · 7 · 4 = 672 such numbers.

Type 2: Numbers with even thousands digits: the thousands digit must be 4 or 6. This will leave5 possible digits for the ones digit, and 8 · 7 choices for the middle two digits. There are2 · 8 · 7 · 5 = 560 such numbers.

Thus, there are 672 + 560 = 1232 such numbers satisfying the conditions.

7. Evaluate1 · 1! + 2 · 2! + 3 · 3! + · · ·+ n · n!

where n ∈ N.

MS Let k = 1 · 1! + 2 · 2! + 3 · 3! + · · ·+ n · n!. Note that

1+k = 1! + 1 · 1!︸︷︷︸2·1!=2!

+2·2!+3·3!+· · ·+n·n! = 2! + 2 · 2!︸︷︷︸3·2!=3!

+3·3!+· · ·+n·n! = · · · = (n+1)·n! = (n+1)!.

Thus k = 1 · 1! + 2 · 2! + 3 · 3! + · · ·+ n · n! = (n + 1)!− 1.

JS1 · 1! + 2 · 2! + 3 · 3! + · · ·+ n · n!

+ 1! + 2! + 3! + · · ·+ n!2! + 3! + 4! + · · ·+ (n + 1)!

− 1! − 2! − 3! − · · ·− n!− 1! + (n + 1)!

8. Evaluate1

(1 + 1)!+

2(2 + 1)!

+ · · ·+ n

(n + 1)!

where n ∈ N.

MS Let k =1

(1 + 1)!+

2(2 + 1)!

+ · · ·+ n

(n + 1)!. Note that

1(1 + 1)!

+2

(2 + 1)!+ · · ·+ n

(n + 1)!

+1

(1 + 1)!+

1(2 + 1)!

+ · · ·+ 1(n + 1)!

2(1 + 1)!

+3

(2 + 1)!+ · · ·+ n + 1

(n + 1)!=

11!

+12!

+ · · ·+ 1n!

Thus k+12!

+13!

+ · · ·+ 1(n + 1)!

=11!

+12!

+ · · ·+ 1n!

and k =12!

+23!

+ · · ·+ n

(n + 1)!= 1− 1

(n + 1)!.

JS Prove, by induction,1

(1 + 1)!+

2(1 + 1)!

+ · · ·+ k − 1k!

= 1− 1k!

:

for k = 2:12!

=12

= 1− 12!

.�

induction: Assume that the formula holds for k = n, then

1(1 + 1)!

+2

(1 + 1)!+ · · ·+ n− 1

n!︸︷︷︸+n

(n + 1)!= 1− 1

n!+

n

(n + 1)!=

(n + 1)!− (n + 1) + n

(n + 1)!

=(n + 1)!− 1

(n + 1)!= 1− 1

(n + 1)!.�

Exercises 1 of 32


9. Prove that for each n ∈ N,(n + 1)(n + 2) · · · (2n)

is divisible by 2n.

10. Find the number of common positive divisors of 1040 and 2030.

JS Since 1040 = 240 · 540 and 2030 = 260 · 530, their common positive divisors are of the form2a · 5b, where 0 ≤ a ≤ 40 and 0 ≤ b ≤ 30. Thus there is a bijection between the common positivedivisors of the numbers and the coordinate pairs (a, b) satisfying the given conditions; both setshave 41 · 31 = 1271 elements.

11. In each of the following, find the number of positive divisors of n (inclusive of n) which are mul-tiples of 3:

(i) n = 210;JS n = 3 · 7 · 5 · 2, thus the each positive divisor of n which is a multiple of 3 is the product of3 and a subset of the set {2, 5, 7}; thus, there are

(30

)+(31

)+(32

)+(33

)= 8 positive divisors of

210 that are multiples of 3.

(ii) n = 630;JS n = 3 · 7 · 5 · 3 · 2, thus the each positive divisor of n which is a multiple of 3 is the productof 3 and a subset of the set {2, 3, 5, 7}; thus, there are

(40

)+(41

)+(42

)+(43

)+(44

)= 16 positive

divisors of 630 that are multiples of 3.

(iii) n = 1512000.JS n = 3 ·7 ·52 ·32 ·25, thus the each positive divisor of n which is a multiple of 3 is the productof 3 and a multi-subset of the multi-set {5 · 2, 2 · 3, 2 · 5, 7}; thus, there are 6 · 3 · 3 · 2 = 108positive divisors of 1512000 that are multiples of 3.

12. Show that for any n ∈ N, the number of positive divisors of n2 is always odd.

KM Let n ∈ N with prime factorization n = p1k1p2

k2 · · · prkr . Then, n2 = p1

2k1p22k2 · · · pr

2kr . Thenumber of positive divisors of n2 is

∏ri=1(2ki+1) which is odd since (2ki+1) is odd and the product

of odd numbers is odd.

13. Show that the number of positive divisors of “111 . . . 1︸︷︷︸1992

” is even.

TT, JA Using some basic divisibility rules, we know that 111 . . . 1111︸︷︷︸1992

is divisible by 3 but not 9.

Thus, we can express 111 . . . 1111︸︷︷︸1992

as

111 . . . 1111︸︷︷︸1992

= 3r.

Let the prime factorization of r be

r = pa11 pa2

2 · · · pak

k where a1, a2, . . . , ak, k ∈ N,

p1, p2, . . . , pk are prime numbers

Since 3r is not divisible by 9, r will not be divisible by 3. Hence, none of p1, p2, . . . , pk will actuallybe a 3.

The prime factorization of 111 . . . 1111︸︷︷︸1992

is simply given by

111 . . . 1111︸︷︷︸1992

= 31pa11 pa2

2 · · · pak

k .

And thus the number of positive divisors that it has will be

(1 + 1)(a1 + 1)(a2 + 1) · · · (ak + 1).

Exercises 1 of 32


We need not concern ourselves with the actual prime factors and their corresponding exponents,because as seen in the expression above, the product is always even.

Hence, the number of positive divisors for 111 . . . 1111︸︷︷︸1992

is always even.

14. Let n, r ∈ N with r ≤ n. Prove each of the following identities:

(i) Pnr = nPn−1

r−1 ,

JS Pnr =

n!(n− r)!

= n(n− 1)!

[(n− 1)− (r − 1)]!= nPn−1

r−1 .

(ii) Pnr = (n− r + 1)Pn

r−1,

JS Pnr =

n!(n− r)!

=n− r + 1n− r + 1

n!(n− r)!

= (n− r + 1)n!

[n− (r − 1)]!= (n− r + 1)Pn

r−1.

(iii) Pnr =

n

n− rPn−1

r , where r < n,

JS Pnr =

n!(n− r)!

=n

n− r

(n− 1)![(n− 1)− r]!

=n

n− rPn−1

r .

(iv) Pn+1r = Pn

r + rPnr−1,

JS

Pnr + rPn

r−1 =n!

(n− r)!+ r

n![n− (r − 1)]!

=n! · (n− r + 1) + r · n!

(n− r + 1)!

=n! · n− n! · r + n! + r · n!

(n− r + 1)!=

n! · (n + 1)[(n + 1)− r]!

=n! · (n + 1)

[(n + 1)− r]!=

(n + 1)![(n + 1)− r]!

= Pn+1r .

JS Pn+1r is the number of arrangements of r distinct objects taken from n + 1 objects. This

number of arrangements may also be counted by considering two cases:Case 1: The (n+1)th object is not among the r distinct objects arranged. Therefore, the r objects

must be from the n other objects, which has Pnr arrangements.

Case 2: The (n + 1)th object is among the r distinct objects arranged. The (n + 1)th object canbe placed in any of the r positions in the arrangement, and the remaining r − 1 distinctobjects are taken from the n remaining objects. There are rPn

r−1 such arrangements.By A.P., the sum Pn

r + rPnr−1 is also the number of arrangements of r distinct objects taken

from n + 1 objects, and therefore Pn+1r = Pn

r + rPnr−1.

(v) Pn+1r = r! + r(Pn

r−1 + Pn−1r−1 + · · ·+ Pr

r−1).JS Using (iv),

Pn+1r = Pn

r + rPnr−1 =⇒ Pn+1

r − Pnr = rPn

r−1

Pnr = Pn−1

r + rPn−1r−1 =⇒ Pn

r − Pn−1r = rPn−1

r−1

Pn−1r = Pn−2

r + rPn−2r−1 =⇒ Pn−1

r − Pn−2r = rPn−2

r−1...

...Pr+1

r = Prr + rPr

r−1 =⇒ Pr+1r − Pr

r = rPrr−1

Pn+1r = r! + r(Pn

r−1 + Pn−1r−1 + · · ·+ Pr

r−1) ⇐= Pn+1r − Pr

r = rPnr−1 + rPn−1

r−1 + · · ·+ rPrr−1

15. In a group of 15 students, 5 of them are female. If exactly 3 female students are to be students areto be selected, in how many ways can 9 students be chosen from the group

(i) to form a committee?JA, JS From the given, we have 5 female students, thus, we have 10 male students, i.e.,15− 5 = 10. Now, we have the following restrictions: that are exactly 3 female students are tobe selected. So, (

53

). (15a)

Exercises 1 of 32


Since we need only exactly 3 students from the group of five female students and knowingthat a committee is composed of 9 students, we have to choose 6 male students from theremaining 10 male students. For that case we have,(

106

). (15b)

Hence, by M.P., multiplying (15a) and (15b) above, there are(53

)(106

)= 2, 100

ways of choosing 9 students from to group of 15 to form a committee.

(ii) to take up 9 different posts in a committee?JA, JS To answer the second question: There are 9! ways for each students to take updifferent posts in a committee since every students in the committee could possibly take upthese 9 position. Thus, by M.P., we have(

53

)(106

)9!.

16. Ten chairs have been arranged in a row. Seven students are to be seated in seven of them so thatno two students share a common chair. Find the number of ways this can be done if no two emptychairs are adjacent?

JS There are 7! ways that the seven students can be arranged in a selection of seven chairs. Atmost one empty chair can be placed between any two chairs that seat students, and there may bean empty chair at the ends of the row, so there are eight possible places to have the three emptychairs, thus

(83

)ways to select the positions of the empty chairs in relation to the chairs that seat

students. By M.P., the number of ways that the students can be seated is 7! ·(83

).

17. Eight boxes are arranged in a row. In how many ways can five distinct balls be put into the boxesif each box can hold at most one ball and no two boxes without balls are adjacent?

JS There are H83 =

(8−3+1

3

)ways to select three empty boxes among the eight boxes so that no

two empty boxes are next to each other, and 5! ways to arrange the balls among the non-emptyboxes. By M.P., the number of ways to place the balls in the boxes are 5! ·

(63

).

18. A group of 20 students, including 3 particular girls and 4 particular boys, are to be lined up in tworows with 10 students each. In how many ways can this be done if the 3 particular girls must bein the front row while the 4 particular boys be in the back?

JS There are 10! ways to arrange the students in the front row, 10! ways to arrange the students inthe back row, and

(136

)ways to choose from the 20− 3− 4 students (other than the particular boys

and girls) six other students to accompany the four boys in the back row. By M.P., the number ofways to line up the students is 10! · 10! ·

(136

).

19. In how many ways can 7 boys and 2 girls be lined up in a row such that the girls must be separatedby exactly 3 boys?

JS There are 2! to arrange the two girls,(73

)ways of choosing the three boys who will be between

the girls from among the seven boys, 3! ways to arrange the chosen boys between the girls, and(4 + 1)! ways to arrange the four remaining boys and the block of five people that have the girls atthe ends. By M.P., the number of ways to line up the boys and girls is 2! ·

(73

)· 3! · 5!.

20. In a group of 15 students, 3 of them are female. If at least one female student is to be selected, inhow many ways can 7 students be chosen from the group

(i) to form a committee?LL, FO There are

(31

)(125

)ways to form the committee with exactly 1 female; there are

(32

)(125

)ways to form the committee with only two females; and there are

(33

)(124

)ways to form the

committee with the three females. By AP, the total number of ways to form the committee with

Exercises 1 of 32


least one female is(31

)(126

)+(32

)(125

)+(33

)(124

)= 5, 643.

LL, FO There are(157

)ways to form the committee of 7 students with no restrictions and

(127

)ways to form the committee with no females. Thus the number of ways to form the committeewith at least 1 female is

(157

)−(127

).

(ii) to take up 7 different posts in a committee?LL, FO Since there are 7! ways for each student to take up different posts in a committeeand every student in the committee could possibly take the 7 positions. Thus, by MP, we have7![(

157

)−(127

)]= (5040)(5643) = 28, 440, 720.

21. Find the number of (m + n)-digit binary sequences with m 0s and n 1s such that no two 1s areadjacent, where n ≤ m + 1.

22. Two sets of parallel lines with p and q lines each are shown in the following diagram:

p

��

��

��

��

��

��

��

︸︷︷︸q

Find the number of parallelograms formed by the lines.

MS Since a parallelogram is formed by two intersecting pairs of parallel lines, choosing two linesamongst the p horizontal lines and two lines amongst the q diagonal lines creates a parallelogram,and each parallelogram in the figure is created by a pair amongst the p horizontal lines and a pairamongst the q diagonal lines, the number of parallelograms in the diagram is equal to the numberof ways that a pair of horizontal lines and a pair of diagonal lines can be selected, which is

(p2

)(q2

).

23. There are 10 girls and 15 boys in a junior class, and 4 girls and 10 boys in a senior class. A commit-tee of 7 members is to be formed from these 2 classes. Find the number of ways this can be done ifthe committee must have exactly 4 senior students and exactly 5 boys.

FO Restriction:

1) 7 members of the committee form 2 classes,

2) exactly 4 senior students,

3) exactly 5 boys.

Consider the ff. cases:

Case 1: Since we only need 7 members of the committee from 2 classes having exactly 4 seniorstudents and exactly 5 boys, we choose 4 boys from the senior class thus we write

(104

).

Because the committee must have exactly 5 boys, we have to pick 7 boys from the juniorclass. We cannot pick another boy from the senior class since we are restricted that exactly4 senior students must be in the committee and it must have exactly 5 boys. Thus we choose1 boy from the junior class having 15 boys and write

(151

). Now, we need to choose 2 female

students from the junior class to complete the committee, we have(102

).

By (MP),(104

)(151

)(102

)= 141, 750 ways of choosing 4 senior boys, 1 boy from the junior class

and 2 girls from the junior class.

Case 2: Choosing3 boys from the senior class

1 girl from the senior class2 boys from the junior class

1 girl from the junior class

By (MP),(103

)(41

)(152

)(101

)= 504, 000 ways.

Exercises 1 of 32


Case 3: Choosing2 boys from the senior class

2 girl from the senior class3 boys from the junior class

By (MP),(102

)(42

)(153

)= 122, 850 ways.

Therefore, by (AP), there are 141, 750 + 504, 000 + 122, 850 = 768, 600 ways a committee of 7members formed from 2 classes having exactly 4 senior students and exactly 5 boys.

24. A box contains 7 identical white balls and 5 identical black balls. They are to be drawn randomly,one at a time without replacement, until the box is empty. Find the probability that the 6th balldrawn is white, while before that exactly 3 black balls are drawn.

PB (52

)(62

)(125

) =25132

.

Note that the first factor in the numerator is the number of ways of arranging the first five balls,three of which are black and two are white. The second factor in the numerator is the number ofways of arranging the last six balls four of which are black and two are white. Also, the sixth ball isfixed to be a white ball. The denominator is just the total number of arranging twelve balls, sevenof which are black and five are white.

25. In each of the following cases, find the number of shortest routes from O to P in the street networkshown below:

rO

rP

rA rBrC

(i) The routes must pass through the junction A;PB We first count the number of ways to go from point O to point A, then from point A to

point P . This is given by:(

52

)(83

)= 560.

(ii) The routes must pass through the street AB;PB We first count the number of ways to go from point O to point A, then from point B to

point P . This is given by:(

52

)(73

)= 350.

(iii) The routes must pass through junctions A and C;PB We first count the number of ways to go from point O to point A, then from point A to

point C, then finally from point C to point P . This is given by:(

52

)(41

)(42

)= 240.

(iv) The street AB is closed.PB We first count the number of ways to go from point O to point P , then we subtract the

number of ways we pass through AB. This is given by:(

135

)−(

52

)(73

)= 937.

26. Find the number of ways of forming a group of 2k people from n couples, where k, n ∈ N with2k ≤ n, in each of the following cases:

(i) There are k couples in such a group;

(ii) No couples are included in such a group;

(iii) At least one couple is included in such a group;

(iv) Exactly two couples are included in such a group.

Exercises 1 of 32


27. Let S = {1, 2, . . . , n + 1} where n ≥ 2, and let

T = {(x, y, z) ∈ S3 | x < z and y < z}.

Show by counting |T | in two two different ways that

n∑k=1

k2 = |T | =(

n + 12

)+ 2(

n + 13

).

KM

Method 1: If z ≥ 2, then there are z − 1 choices for both x and y. Thus, |T | =∑n+1

i=2 (i− 1)2 =∑n

k=1 k2.Method 2: Consider 2 disjoint cases.

Case 1: x = y. Then, there are(n+1

2

)such triples. Choose 2 numbers from S, the larger one is z,

and the smaller one is x and y.Case 2: x 6= y. Then, there are 2

(n+1

3

)such triples. First, choose 3 numbers from S, the largest

one is z. We multiply by 2 because there are two possible positions for the remaining twonumbers from S (after having set z).

Thus, |T | =(n+1

2

)+ 2(n+1

3

).

28. Consider the following set of points on the xy-plane:

A = {(a, b) | a, b ∈ Z, 0 ≤ a ≤ 9 and 0 ≤ b ≤ 5}.

Find

PB Note that we have 10 a coordinates, and 6 b coordinates.

(i) the number of rectangles whose vertices are points in A;PB To form a rectangle, we simply select two points from the 10 a coordinates, and two from

the 6 b coordinates. This is given by:(

102

)(62

)= 675.

(ii) the number of squares whose vertices are points in A.PB We count the number of squares using cases:

1× 1 square: 5 · 9 such squares2× 2 square: 4 · 8 such squares3× 3 square: 3 · 7 such squares4× 4 square: 2 · 6 such squares5× 5 square: 1 · 5 such squares

Hence, there are 115 squares.

29. Fifteen points P1, P2, . . . , P15 are drawn in the plane in such a way that besides P1, P2, P3, P4, P5

which are drawn collinear, no other 3 points are collinear. Find

(i) the number of straight lines which pass through at least 2 of the fifteen points;PB We basically count the number of ways of connecting any two points. Now, since fivepoints are collinear, then we subtract the number of ways of connecting any two of these fivepoints, and add 1 for the line formed by these five collinear points. This is given by:(

152

)−(

52

)+ 1 = 96.

(ii) the number of triangles whose vertices are 3 of the 15 points.PB We basically count the number of ways of connecting any three points to form a triangle.Now, since five points are collinear, then we subtract the number of ways of connecting anythree of these five points since those three will not form a triangle. This is given by:(

153

)−(

53

)= 445.

Exercises 1 of 32


30. In each of the following 6-digit natural numbers:

333333, 225522, 118818, 707099,

every digit in the number appears at least twice. Find the number of such 6-digit natural numbers.

31. In each of the following 7-digit natural numbers:

1001011, 5550000, 3838383, 7777777,

every digit in the number appears at least 3 times. Find the number of such 7-digit natural num-bers.

MS A 7-digit must begin with a nonzero digit, giving nine possible options. In relation to the firstdigit, there are three cases to consider:

Case 1: The first digit is repeated twice in the next six digits. The four remaining digits will be thesame, so the possibilities of this case are determined by the location of the two repetitions ofthe first digit, and the choice of the remaining digit. Hence, this case has 9 ·

(62

)· 9 elements.

Case 2: The first digit is repeated thrice in the next six digits. The three remaining digits will be thesame, so the possibilities of this case are determined by the location of the three repetitionsof the first digit, and the choice of the remaining digit. Hence, this case has 9 ·

(63

)·9 elements.

Case 3: The first digit is repeated in the next six digits. This case has 9 elements.

So there are a total of 9 ·(62

)· 9 + 9 ·

(63

)· 9 + 9 = 2844 such 7-digit natural numbers.

32. Let X = {1, 2, 3, . . . , 1000}. Find the number of 2-element subsets {a, b} of X such that the producta · b is divisible by 5.

PB Any product of two numbers is divisible by 5 if at least one of the two numbers is a multipleof 5. Thus the product of (5k)(a) is divisible by 5 for any natural numbers a and k. Let us nowanswer the problem. The 2-element subsets we are looking for is in the form {5k, a} (order doesnot matter). Thus, the smallest value of k is 1 and the largest is k = 200. Thus we consider 200cases.

Case 1. Let k = 1, thus {5k, a} = {5, a}. There are 999 subsets of this form. (All one thousandelements except 5).

Case 2. Let k = 2, thus {5k, a} = {10, a}. Again, there are 999 subsets of this form. However, thesubset {10, 5} is already counted in the first case. Thus we only count the subsets of this formthat are not in the previous case. There are 998 such subsets. (All one thousand elementsexcept 5 and 10).

...

Case 200. Let k = 200 thus {5k, a} = {1000, a}. There are 800 subsets that are distinct from the previouscases. (All one thousand elements except 5, 10, 15, . . . , 990, 995 and 1000).

Thus there are 999 + 998 + 997 + · · ·+ 801 + 800 = 179900 2-element subsets of X whose productof the elements is divisible by 5.

TT We shall partition the set X into two sets: X1 = {5, 10, 15, . . . , 995, 1000} and X2 = {1, 2, 3, 4, 6,. . . , 997, 998, 999}. Note that all elements in X1 are divisible by 5 while those in X2 are not. Fur-thermore, |X1| = 200 and |X2| = 800. To get two element subsets of X such that the product itselements is divisible by 5, we either get two elements from set X1, which can be done in

(2002

)ways, or get one element each from set X1 and X2, which can be done in

(2001

)(8001

)ways. Thus

by AP, the total number of such subsets is(2002

)+(2001

)(8001

)= 179900.

33. Consider the following set of points in the xy-plane:

A = {(a, b) | a, b ∈ Z and |a|+ |b| ≤ 2}.

Find

Exercises 1 of 32


(i) |A|;(ii) the number of straight lines which pass through at least 2 points in A; and

(iii) the number of triangles whose vertices are points in A.

34. Let P be a convex n-gon, where n ≥ 6. Find the number of triangles formed by any 3 vertices of Pthat are pairwise nonadjacent in P .

35. 6 boys and 5 girls are to be seated around a table. Find the number of ways that this can be donein each of the following cases:

(i) There are no restrictions;PB (11− 1)! = 10! ways.

(ii) No 2 girls are adjacent;PB We first arrange the 6 boys around the table, then insert the girls in between the boys.This guarantees that no two girls will be seated next to each other. Hence we have: 5!

(65

)5!

ways.

(iii) All girls form a single block;PB We first chunk all the girls and treat them as one “object”. We now have a total of 7“objects”, which we have to arrange around the table. We then permute the girls inside thechunk. Hence we have: 6!5! ways.

(iv) A particular girl G is adjacent to two particular boys B1 and B2.PB We first chunk [B1GB2], so now we have a total of 9 “objects”, which we have to arrangearound the table. Then we permute B1 and B2. Hence we have: 8!2! ways.

36. Show that the number of r-circular permutations of n distinct objects, where 1 ≤ r ≤ n, is givenby n!

(n−r)!·r .

37. Let k, n ∈ N. Show that the number of ways to seat kn people around k distinct tables such thatthere are n people in each table is given by (kn)!

nk .

38. Let r ∈ N such that1(9r

) − 1(10r

) =11

6(11r

) .Find the value of r.

FO

19!

r!(9−r)!

− 110!

r!(10−r)!

=11

6( 10!r!(11−n)! )

⇐⇒ r!(9− r)!9!

− r!(10− r)!10!

=(11)(r!)(11− r)!

6(11!)

⇐⇒ r!(9− r)!9!

(1− (10− r)

10

)=

(11)(r!)(11− r)!6(11!)

⇐⇒ 9!r!(9− r)!

(r!(9− r)!

9!

( r

10

))=[(11)(r!)(11− r)(10− r)(9− r)!

6(11)(10)(9!)

]9!(10)

r!(9− r)!

⇐⇒ r =(11− r)(10− r)

6⇐⇒ 6r = 110− 21r + r2 ⇐⇒ r2 − 27r + 110 = 0

⇐⇒ (r − 22)(r − 5) = 0 ⇐⇒ r = 22 or r = 5

But it is impossible for(nr

)where r > n — by definition, r ≤ n.

Thus, r = 5.

39. Prove each of the following identities:

(a)(

n

r

)=

n

r

(n− 1r − 1

), where n ≥ r ≥ 1;

LLn

r

(n− 1r − 1

)=

n

r

[(n− 1)!

(r − 1)!(n− 1− r + 1)!

]=

n!r!(n− r)!

=(

n

r

).

Exercises 1 of 32


(b)(

n

r

)=

n− r + 1r

(n

r − 1

), where n ≥ r ≥ 1;

LL

n− r + 1r

(n

r − 1

)=

n− r + 1r

[n!

(r − 1)!(n− r + 1)!

]=

n!(n− r + 1)r!(n− r + 1)!

=n!

r!(n− r)!=(

n

r

).

(c)(

n

r

)=

n

n− r

(n− 1

r

), where n > r ≥ 0;

LLn

n− r

(n− 1

r

)=

n

n− r

[(n− 1)!

r!(n− 1− r)!

]=

n!r!(n− r)!

=(

n

r

).

(d)(

n

m

)(m

r

)=(

n

r

)(n− r

m− r

), where n ≥ m ≥ r ≥ 0.

LL(n

r

)(n− r

m− r

)=

n!r!(n− r)!

[(n− r)!

(m− r)!(n− r −m + r)!

]=

n!(n− r)!r!(n− r)!(m− r)!(n−m)!

· m!m!

=n!

m!(n−m)!· m!r!(m− r)!

=(

n

m

)(m

r

).

40. Prove the identity(nr

)=(

nn−r

)by using the bijection principle.

41. Let X = {1, 2, . . . , n}, A = {A ⊂ X | n /∈ A}, and B = {A ⊂ X | n ∈ A}. Show that |A| = |B| byusing the bijection principle.

MS For every A ∈ A, X \A ∈ B; for every B ∈ B, X \B ∈ A. Therefore, |A| = |B|.

42. Let r, n ∈ N. Show that the product

(n + 1)(n + 2) · · · (n + r)

of r consecutive positive integers is divisible by r!.

MS (n + 1)(n + 2) · · · (n + r) = Pn+rr = r!

(n+r

r

). Hence, r! | (n + 1)(n + 2) · · · (n + r).

43. Let A be a set of kn elements, where k, n ∈ N. A k-grouping of A is a partition of A into k-elementsubsets. Find the number of different k-groupings of A.

KM For the first k-subset,(knk

). For the 2nd k-subset,

(kn−k

k

). For the 3rd k-subset,

(kn−2k

k

). And

for the nth k-subset,(kn−(n−1)k

k

)=(kk

). Hence, the number of different k-groupings would seem

to be(knk

)(kn−k

k

)(kn−2k

k

)· · ·(kk

). But we still need to divide by the number of ways to arrange the

n k-subsets since the the ordering of the k-subsets doesn’t matter. Therefore, the final answer is(knk

)(kn−k

k

)(kn−2k

k

)· · ·(kk

)n!

.

44. Twenty-five of King Arthur’s knights are seated at their customary round table. Three of them arechosen — all choices of three being equally likely — and are sent off to slay a troublesome dragon.Let P be the probability that at least two of the three had been sitting next to each other. If P iswritten as a fraction in lowest terms, what is the sum of the numerator and denominator?

KM Number in sample space =(253

)= 2300. How many ways can we choose 3 knights such that

at least two of the three had been sitting next to each other (event E)?

Case 1: There are exactly 2 knights sitting next to each other. There are 25 choices for the first knight,there are 2 ways to choose a seatmate (his left or right), and there are 21 ways to selectanother knight which is not adjacent to either of the knights already chosen. Divide by 2!since we counted twice the possibility that the seatmate was chosen first and the original waschosen next. Ans: (25)(2)(21)

2! = 525.

Case 2: There are exaclty 3 knights sitting next to each other. Ans: There are 25 such consecutive”triples”.

Exercises 1 of 32


Therefore, P (E) =525 + 25

2300=

5502300

=1146

and the sum of the numerator and the denominator is57.

45. One commercially available ten-button lock may be opened by depressing — in any order — thecorrect five buttons. The sample shown below has {1, 2, 3, 6, 9} as its correct combination. Supposethat these locks are redesigned so that sets of as many as nine buttons or as few as one button couldserve as combinations. How many additional combinations would this allow?

12345

678910

a a aa a aa a a

a a a

a a a

MS Currently, the locks require five buttons to unlock, which indicates that there are(105

)com-

binations. If, after redesigning, as few as one button to as many as nine buttons can be usedas a combination, only two combinations will not be viable: the combination where no button isdepressed

(100

), and the combinations where all buttons are depressed

(1010

). Hence, the number

of combinations added by the redesign is(101

)+(102

)+(103

)+(104

)+(106

)+(107

)+(108

)+(109

)= 770.

46. In a shooting match, eight clay targets are arranged in two hanging columns of three each and onecolumn of two, as pictured. A marksman is to break all eight targets accoring to the followingrules: (1) The marksman first chooses a column from which a target is to be broken. (2) Themarksman must then break the lowest remaining unbroken target in the chosen column. If theserules are followed, in how many different orders can the eight targets be broken?

i i i

i i i

i i

TT Consider a string of letters that will be composed of three As, two Bs and three Cs. The numberof ways in which we can construct such a string is equal to 8!

3!·2!·3! .

Now let’s note a particular string, say ABACCABC. If we are to relate this to the given problem, wecan treat the three As as the clay targets on the first column, the two Bs as the clay targets on thesecond column, and the three Cs as the clay targets on the third column.

Considering the given arrangement of letters in the string, this can be interpreted as:

(1) on the first shot, the first (or bottom) clay target on the first column was hit;(2) on the second shot, the first (or bottom) clay target of the second column was hit;(3) on the third shot, the second (or middle) clay target of the first column was hit;(4) on the fourth shot, the first (or bottom) clay target of the third column was hit;(5) on the fifth shot, the second (or middle) clay target of the third column was hit;(6) on the sixth shot, the third (or top) clay target of the first column was hit;(7) on the seventh shot, the second (or top) clay target of the second column was hit;(8) and finally on the last shot, the remaining clay target on the third column was hit.

Note that the interpretation is still consistent with the rules stated in the problem.

Therefore, we can say that there is a bijection between the number of ways in which we canarrange three As, two Bs and three Cs to form a string, and the number of ways in which a shootercan break all the 8 clay targets according to the rules.

Exercises 1 of 32


Thus the number of ways a shooter can accomplish the task is equal to 8!3!·2!·3! = 560.

47. Using the numbers 1, 2, 3, 4, 5 we can form 5!(= 120) 5-digit numbers in which the 5 digits are alldistinct. If these numbers are listed in increasing order:

123451st

, 123542nd

, 124353rd

, . . . , 54321120th

,

find (i) the position of the number 35421; (ii) the 100th number in the list.

FO

12345 12543 13524 14352 15324 21354 2314512354 13245 13542 14523 15342 21435 2315412435 13254 14235 14532 15423 21453 2341512453 13425 14253 15234 15432 21534 2345112534 13452 14325 15243 21345 21543 23514

23541 31524 35241 43215 5234124135 31542 35412 43251 5231424153 32145 35421 43512 5241324315 32154 41235 43521 5243124351 32415 41253 45123 5312424513 34251 41325 45132 5314224531 32514 41352 45213 5321425134 32541 41523 45231 5324125143 34125 41532 45312 5412325314 34152 42135 45321 5413225341 34215 42153 51234 5421325413 34251 42315 51243 5423125431 34512 42351 51324 5431231245 34521 42513 51342 5432131254 35124 42531 5142331425 35142 43125 5213431452 35214 43152 52143

(i) The position of the number 35421 is in the 72nd position. (ii) the 100th number in the list is thenumber 51342.

48. The P43(= 24) 3-permutations of the set {1, 2, 3, 4} can be arranged in the following way, called the

lexicographic ordering:

123, 124, 132, 134, 142, 143, 213, 214, 231, 234,241, 243, 312, . . . , 431, 432.

Thus the 3-permutations “132” and “214” appear at the 3rd and 8th positions of the orderingrespectively. There are P9

4(= 3024) 4-permutations of the set {1, 2, . . . , 9}. What are the positionsof the 4-permutations “4567” and “5182” in the corresponding lexicographic ordering of the 4-permutations of the set {1, 2, . . . , 9}?

KM Let us consider first 4567. We will find all the permutations less than 4567. There are(3)(8)(7)(6) = 1008 numbers less than 4000. (The thousands place can only come from {1, 2, 3}.)There are (1)(3)(7)(6) = 126 numbers greater than 4000 but less than 4500. (The hundreds canonly be {1, 2, 3}. The thousands place is already 4.) There are (1)(1)(3)(6) = 18 numbers greaterthan 4500 but less than 4560. (The tens can only be {1, 2, 3}.) Lastly, there are (1)(1)(1)(3) =3 numbers greater than 4560 but less than 4567. (The ones can come from {1, 2, 3}.) Thus,the total number of permutations less than 4567 totals to 1155, implying that 4567 is the 1156thpermutation in lexicographic ordering.�Finding the position of 5182 is done similarly. There are (4)(8)(7)(6) = 1344 numbers less than5000. (The thousands can be {1, 2, 3, 4}.) There are no permutations greater than 5000 but lessthan 5100. There are (1)(1)(5)(6) = 30 numbers greater than 5100 and less than 5180. (The tenscan come from {2, 3, 4, 6, 7} and the ones can come from those remaining from the choices for thetens (4), along with 8 and 9. This gives 6 choices for the ones.) Lastly, there are no permutationsgreater than 5180 but less than 5182. Thus, the total number of permutations less than 5182 totalsto 1374, implying that 5182 is the 1375th permutation in lexicographic ordering.�

Exercises 1 of 32


49. The(53

)(= 10) 3-element subsets of the set {1, 2, 3, 4, 5} can be arranged in the following way, called

the lexicographic ordering:

{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5},{2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}.

Thus the subset {1, 3, 5} appears in the 5th position of the ordering. There are(104

)4-element

subsets of the set {1, 2, . . . , 10}. What are the positions of the subsets {3, 4, 5, 6} and {3, 5, 7, 9} inthe corresponding lexicographic ordering of the 4-element subsets of {1, 2, . . . , 10}?

PB Notice that there are six 3-element subsets of the form {1, a, b}, where a and b are distinctelements of the set {2, 3, 4, 5}. Since the subset {2, 3, 4} appears right next to these six subsets,then it should appear in the 7th position. Keeping this in mind, it would now be easy for us todetermine the position of a given subset.

There are(93

)4-element subsets of the form {1, a, b, c}, where a, b and c are distinct elements of

the set {2, 3, . . . , 10}; and there are(83

)4-element subsets of the form {2, d, e, f}, where d, e and f

are distinct elements of the set {3, 4, . . . , 10}. Also, the subset {3, 4, 5, 6} is the first subset startingwith 3. Hence the subset {3, 4, 5, 6} appears in the

[(93

)+(83

)+ 1 = 141

]th position.

More so, there are(62

)4-element subsets of the form {3, 4, g, h}, where g and h are distinct ele-

ments of the set {5, 6, . . . , 10}; and(41

)4-element subsets of the form {3, 5, 6, i}, where i is an ele-

ment of the set {7, 8, 9, 10}. Since the subset {3, 5, 7, 9} is the second subset of the form {3, 5, 7, j},where j is an element of the set {8, 9, 10}, then it should appear in the position numbered(

93

)+(

83

)+(

62

)+(

41

)+ 2 = 161.

50. Six scientists are working on a secret project, They wish to lock up the documents in a cabinet sothat the cabinet can be opened when and only when three or more of the scientists are present.What is the smallest number of locks needed? What is the smallest number of keys each scientistmust carry?

51. A 10-storey building is to be painted with some 4 different colors such that each storey is paintedwith one color. It is not necessary that all the 4 colors must be used. How many ways are there topaint the building if

(i) there are no other restrictions?PB Each floor has four choices, hence there are 410 ways to paint the building.

(ii) any 2 adjacent stories must be painted with different colors?PB Suppose we start painting with the ground floor, then initially we have four choices. Butfor the second floor, we only have three choices not including our previous choice of color.Again, for the third floor, we only have three choices, and so on. Thus there are 4 · 39 ways topaint the building.

52. Find the number of all multi-subset of M = {r1 · a1, r2 · a2, . . . , rn · an}.

TT To find the number of all multi-subsets of M = {r1 · a1, r2 · a2, . . . , rn · an}, note that, for everyobject ai, there are (1 + ri) ways of “choosing” it — that is, the multisubset of M may contain noneof object ai, only one of object ai, two of object ai, up untill all of object ai (for a total (1+ri) ways inwhich object ai can be a part of the multi-subset). Thus by (MP), the total number of multi-subsetsof M is equal to (1 + r1)(1 + r2) · · · (1 + rn).

53. Let r, n ∈ N with r ≤ n. A permutation x1x2 · · ·x2n of the set {1, 2, . . . , 2n} is said to have theproperty P (r) if |xi − xi+1| = r for at least one i in {1, 2, . . . , 2n− 1}. Show that, for each n and r,there are more permutations with property P (r) than without.

54. Prove by combinatorial argument that each of the following numbers us always an integer for eachn ∈ N:

TT The following expressions can be seen as the number of ways of arranging nk objects, such

Exercises 1 of 32


that k are of type 1, k are of type 2, . . . , k are of type n. And since these expressions are a resultof counting the number of arrangements, then they must always be an integer.

Note that the number of ways to arrange such nk objects is given by

(kn)!k! · k! · · · k!︸︷︷︸

n times

.

(i)(3n)!2n · 3n

,

TT(3n)!2n · 3n

=(3n)!(3!)n

=(3n)!

3! · 3! · · · 3!︸︷︷︸n times

.

This can viewed as the number of ways of arranging 3n objects, such that 3 are of type 1, 3are of type 2, . . . , 3 are of type n.

(ii)(6n)!

5n · 32n · 24n,

TT(6n)!

5n · 32n · 24n=

(6n)!(6!)n

=(6n)!

6! · 6! · · · 6!︸︷︷︸n times

.

This can viewed as the number of ways of arranging 6n objects, such that 6 are of type 1, 6are of type 2, . . . , 6 are of type n.

(iii)(n2)!(n!)n

,

TT(n2)!(n!)n

=(n2)!

n! · n! · · ·n!︸︷︷︸n times

.

This can viewed as the number of ways of arranging n2 objects, such that n are of type 1, nare of type 2, . . . , n are of type n. Note that the total number of objects, given you have nkinds, each with n objects, is n · n = n2.

(iv)(n!)!

(n!)(n−1)!.

TT(n!)!

(n!)(n−1)!=

[n · (n− 1)!]!n! · n! · · ·n!︸︷︷︸

(n−1)! times

.

This can viewed as the number of ways of arranging n! objects, such that n are of type 1, nare of type 2, . . . , n are of type (n− 1)!. Note that the total number of objects, given you have(n− 1)! kinds, each with n objects, is n · (n− 1)! = n!.

55. Find the number of r-element multi-subsets of the multi-set

M = {1 · a1,∞ · a2,∞ · a3, . . . ,∞ · an}.

TT An r-element multi-subset of the multi-set M = {1 ·a1,∞·a2, . . . ,∞·an}may either (1) containa1, or (2) not contain a1.

Let xi represent the number of object ai in the subset. To determine the number of subsetsin (1), we only need to consider the number of non-negative integer solutions to the equationx1 +x2 + · · ·+xn = r, where x1 = 1. This is just equal to

((r−1)+(n−1)−1

r−1

)=(r+n−3

r−1

). To determine

the number of subsets in (2), we determine the number of non-negative integer solutions to theequation x2 + · · ·+ xn = r. This is just equal to

((r)+(n−1)−1

r

)=(r+n−2

r

).

Hence by (AP), the total number of r-element multi-subsets of M is equal to(r+n−3

r−1

)+(r+n−2

r

).

Exercises 1 of 32


56. Six distinct symbols are transmitted through a communication channel. A total of 18 blanks areto be inserted between the symbols with at least 2 blanks between every pair of symbols. In howmany ways can the symbols and the blanks be arranged?

PB We first permute the six distinct symbols then fix the ten blanks in such a way that exactly twoblanks are between every pair of symbols. This leaves us with eight more blanks to be insertedin five spaces. Therefore the number of ways to arrange the symbols and the blanks is given by:(8+5−15−1

)6! = 356400.

57. In how many ways can the following 11 letters: A, B, C, D, E, F, X, X, X, Y, Y be arranged in a rowso that every Y lies between two Xs (not necessarily adjacent)?

PB We first fix the three Xs. Since we want every Y to be in between two Xs, then we insert the Ysinto the spaces between the Xs. Next, we insert the remaining six letters anywhere between theXs and Ys including the “outside” spaces (space to the left of the leftmost X and to the right of therightmost X). Finally, we permute these six letters. Hence we have:

(2+2−1

2

)(6+6−1

6

)6!.

58. Two n-digit integers (leading zero allowed) are said to be equivalent if one is a permutation of theother. For instance, 10075, 01057 and 00751 are equivalent 5-digit integers.

(i) Find the number of 5-digit integers such that no two are equivalent.TT The number of 5-digit integers such that no two are equivalent is just the same as thenumber of 5-element multi-subsets of M = {∞ · 0,∞ · 1, . . . ,∞ · 9}, or the number of non-negative integer solutions to the equation x0 +x1 + · · ·+x9 = 5, where xi denotes the numberof times digit i was used, which is just equal to

(145

).

(ii) If the digits 5, 7, 9 can appear at most once, how many nonequivalent 5-digit integers arethere?TT If 5, 7, 9 can appear at most once, we shall consider four cases:

(a) none of 5, 7, 9 are includedThe number of such 5-digit integers is equal to the number of non-negative integer solu-tions to the equation

∑9i=0

i 6=5,7,9

xi = 5. This will give us(115

)numbers.

(b) one of the numbers 5, 7, 9 is includedThe number of such 5-digit integers is equal to the number of non-negative integer solu-tions to the equation

( ∑9i=0

i 6=5,7,9

xi

)+ xs = 5, where xs = 1, for s = 5, 7 or 9. This will give

us(31

)(104

)numbers.

(c) two of the numbers 5, 7, 9 are includedThe number of such 5-digit numbers is equal to the number of non-negative integer solu-tions to the equation

( ∑9i=0

i 6=5,7,9

xi

)+ (xs + xt) = 5, where xs = xt = 1, for s, t = 5, 7, 9 and

s > t. This will give us(32

)(93

)numbers.

(d) all of the numbers 5, 7, 9 are includedThe number of such 5-digit numbers is equal to the number of non-negative integer solu-tions to the equation

∑9i=0

i 6=5,7,9

xi = 2. This will give us(82

)numbers.

Hence, the total number of such 5-digit numbers is equal to(115

)+ 3(104

)+ 3(93

)+(82

).

59. How many 10-letter words are there using the letters a, b, c, d, e, f if

(i) there are no restrictions?KM, LL 10-letter word, each letter has 6 choices. Ans: 610.

(ii) each vowel (a and e) appears 3 times and each consonant appears once?

KM Just count all possible arrangements of the word aaaeeebcdf. Ans:10!3!3!

.

LL The number of such 10-letter words is the number of permutations of the multi-set {3 ·a, b, c, d, 3 · e, f}, which is equal to 10!

3!3! .

Exercises 1 of 32


(iii) the letters in the word appear in alphabetical order?KM This is bijective to looking for all nonnegative integer solutions to x1 + x2 + · · ·+ x6 = 10.

Ans: H610 =

(10 + 6− 1

10

)=(

1510

).

(iv) each letter occurs at least once and the letters in the word appear in alphabetical order?KM This is bijective to looking for all the positive integer solutions to x1 + x2 + · · ·+ x6 = 10.

Ans: H64 =

(4 + 6− 1

4

)=(

94

).

60. Let r, n, k ∈ N such that r ≥ nk. Find the number of ways of distributing r identical objects into ndistinct boxes so that each box holds at least k objects.

FOk

1k

2k

3· · · k

n

First put k objects in each box to fulfill the requirements. This can be done in one way. Thendistribute the remaining r − nk objects in the boxes in an arbitrary way. Using (MP) and result incase 2(ii) (

r − nk + n− 1r − nk

)=(

r − n(k − 1)− 1r − nk

);

by identity (1.4.2) this can also be written as(r − n(k − 1)− 1

n− 1

).

61. Find the number of ways of arranging the 9 letters r, s, t, u, v, w, x, y, z in a row so that y alwayslies between x and z (x and y, or y and z need not be adjacent in the row).

PB We first fix x, y and z respectively so that y is always between x and z. We then permutex and z. Next we insert the remaining six letters to the spaces between x, y and z, including the“outside” spaces. Finally we permute these six letters. Hence we have: 2!

(4+6−1

6

)6!.

62. Three girls A, B and C, and nine boys are to be lined up in a row. In how many ways can this bedone if B must lie between A and C, and A, B must be separated by exactly 4 boys?

TT Since B is in between A and C, then the girls can only be arranged in two ways: CBA andABC. Now consider a particular arrangement, ABC.

9 boys must be placed in between/beside the girls; assume for now that every boy is the same.Since there should be exactly 4 boys in between A and B, we are going to have AY Y Y Y BC (Yrepresents a boy). Next we’re going to place the boys to the left of A, between B and C, and tothe right of C. There are 3 spaces where the 5 remaining boys can position themselves in, and thenumber of ways in which they can to that is given by

(5+3−1

5

). Since the 9 boys are different from

one another, by MP we shall have(5+3−1

5

)· 11!.

Since the number of ways of arranging the boys satisfying the given conditions given the initialarrangement of girls ABC is the same as that for CBA, then the total number of arrangements isjust given by

(5+3−1

5

)· 11! · 2 = 42 · 11!.

63. Five girls and eleven boys are to be lined up in a row such that, from left to right, the girls are inthe order G1, G2, G3, G4, G5. In how many ways can this be done if G1 and G2 must be separatedby at least 3 boys, and there is at most one boy between G4 and G5?

64. Given r, n ∈ N with r ≥ n, let L(r, n) denote the number of ways of distributing r distinct objectsinto n identical boxes so that no box is empty and the objects in each box are arranged in a row.Find L(r, n) in terms of r and n.

KM The number of ways to place r identical objects into n distinct boxes is(

r−1n−1

). Multiply by

r! since the objects must be distinct and the arrangement matters. Divide by n! since the boxesmust be identical and hence, its arrangements must be removed from the counting. Therefore,L(r, n) = r!

n!

(r−1n−1

).

Exercises 1 of 32


65. Find the number of integer solutions to the equation:

x1 + x2 + x3 + x4 + x5 + x6 = 60

in each of the following cases:

(i) xi ≥ i− 1 for each i = 1, 2, . . . , 6;PB By subtracting i − 1 to xi for each i = 1, 2, . . . , 6, we have adjusted the lower bound foreach xi to zero instead of i−1. However, for the original equation to remain constant, we alsohave to subtract

∑6i=1(i− 1) from 60. Hence the number of integer solutions to the equation

given above is the same to the number of nonnegative integer solutions to the equation:

y1 + y2 + y3 + y4 + y5 + y6 = 45.

Hence we have(45+6−1

45

)=(505

).

(ii) x1 ≥ 2, x2 ≥ 5, 2 ≤ x3 ≤ 7, x4 ≥ 1, x5 ≥ 3 and x6 ≥ 2.PB Again we have to adjust each lower bound to zero instead of some other number for ourgeneral result to work. However, notice that x3 has an upper bound. Let us ignore this firstfor the meantime. By the same technique shown above, the number of integer solutions tothe equation above (ignoring x3’s upper bound) is the same as the number of nonnegativeinteger solutions to the equation:

y1 + y2 + y3 + y4 + y5 + y6 = 60− 2− 5− 2− 1− 3− 2 = 45.

Hence we have(505

). However, we still have to remove all the cases when x3 ≥ 8 and there

are(39+6−1

39

)=(445

)of them. Hence our final answer is

(505

)−(445

).

66. Find the number of integer solutions to the equation:

x1 + x2 + x3 + x4 = 30

in each of the following cases:

(i) xi ≥ 0 for each i = 1, 2, 3, 4;PB From our general result, the number of nonnegative integer solutions to the equationabove is given by:

(30+4−1

30

)=(333

).

(ii) 2 ≤ x1 ≤ 7 and xi ≥ 0 for each i = 2, 3, 4;PB By the same technique shown in the previous number, we have:

(28+4−1

28

)−(22+4−1

22

)=(

313

)−(253

).

(iii) x1 ≥ −5, x2 ≥ −1, x3 ≥ 1 and x4 ≥ 2.PB Again, the number of integer solutions to this equation is the same as the number ofnonnegative integer solutions to the equation:

y1 + y2 + y3 + y4 = 30 + 5 + 1− 1− 2 = 33.

Hence we have:(33+4−1

33

)=(363

).

67. Find the number of quadruples (w, x, y, z) of nonnegative integers which satisfy the inequality

w + x + y + z ≤ 1992.

TT, JA The given inequality can be converted to an equation by introducing a slack variable s,where s is nonnegative:

w + x + y + z + s = 1992.

The slack variable s captures any excess value not taken up by the original variables. That meanswhen w + x + y + z = 1991, then s = 1; when w + x + y + z = 1987, then s = 5.

Since s is non-negative, it would not be possible for w + x + y + z > 1992, as this will force s tobecome negative. Thus the number of quadruplets (w, x, y, z) satisfying the inequality is the samethe number of nonnegative integer solutions for the equation w + x + y + z + s = 1992. This canbe interpreted as the number of ways of arranging 1992 sticks and 4 “+” signs, is equal to

(1996

4

).

Exercises 1 of 32


68. Find the number of nonnegative integer solutions to the equation:

5x1 + x2 + x3 + x4 = 14.

JA, LL Consider the following cases:

Case 1: What if x1 = 0? Then the equation becomes

x2 + x3 + x4 = 14.

In this case, we have (n + r − 1

r − 1

)=(

14 + 22

)=(

162

).


5(1) + x2 + x3 + x4 = 14,

which is equal tox2 + x3 + x4 = 9,

and by the same reasoning, we have(n + r − 1

r − 1

)=(

9 + 22

)=(

112

).


5(2) + x2 + x3 + x4 = 14 ⇔ x2 + x3 + x4 = 4.

In this case, we have (n + r − 1

r − 1

)=(

4 + 22

)=(

62

).

So, by A.P., we have, (162

)+(

112

)+(

62

)= 120 + 55 + 15 = 190.


rx1 + x2 + · · ·+ xn = kr,

where k, r, n ∈ N.

JA, LL Consider the following cases:


x2 + · · ·+ xn = kr.

In this case, we have n− 2 pluses and so(kr + n− 2

n− 2

).


r(1) + x2 + · · ·+ xn = kr

which is equal tox2 + · · ·+ xn = kr − r = r(k − 1)

and by the same reasoning, we have(r(k − 1) + n− 2

n− 2

).

Exercises 1 of 32


We do this inductively until we reach this case:

Case k + 1: What if x1 = k? Then the equation becomes

r(k) + x2 + · · ·+ xn = kr

which is equal tox2 + · · ·+ xn = kr − kr = r(k − k)

and by the same reasoning, we have(r(k − k) + n− 2

n− 2

)=(

n− 2n− 2

).

Getting a general form of this (after some thought) we have,

k∑i=0

(r(k − i) + n− 2

n− 2

).


3x1 + 5x2 + x3 + x4 = 10.

PB Since each xi ≥ 0 for i = 1, 2, 3, 4, then 0 ≤ x1 ≤ 3 and 0 ≤ x2 ≤ 2. We now consider 4 ·3 = 12cases:

(x1, x2) # of nonnegative integer solutions (x1, x2) # of nonnegative integer solutions(0, 0)

(10+2−1

10

)=(111

)= 11 (1, 1)

(2+2−1

2

)=(31

)= 3

(1, 0)(7+2−1

7

)=(81

)= 8 (1, 2) ∅ → 0

(2, 0)(4+2−1

4

)=(51

)= 5 (2, 1) ∅ → 0

(3, 0)(1+2−1

1

)=(21

)= 2 (2, 2) ∅ → 0

(0, 1)(5+2−1

5

)=(61

)= 6 (3, 1) ∅ → 0

(0, 2)(0+2−1

0

)=(11

)= 1 (3, 2) ∅ → 0

Hence there are 36 nonnegative integer solutions to the equation above.

71. Find the number of positive integer solutions to the equation:

(x1 + x2 + x3)(y1 + y2 + y3 + y4) = 77.

JA Consider the following cases:

Case 1: If x1 +x2 +x3 = 7 and y1 +y2 +y3 +y4 = 11, then the number of nonnegative integer solutionsto the equation x1 + x2 + x3 = 7 is given by(

r + n− 1r

)=(

7 + 3− 17

)=(

97

)=(

92

),

and to the equation y1 + y2 + y3 + y4 = 11 is given by(r + n− 1

r

)=(

11 + 4− 111

)=(

1411

)=(

143

).

By M.P., the desired number is given by (92

)(143

).

Exercises 1 of 32


Case 2: If x1 +x2 +x3 = 11 and y1 +y2 +y3 +y4 = 7, then the number of nonnegative integer solutionsto the equation y1 + y2 + y3 + y4 = 7 is given by(

r + n− 1r

)=(

7 + 4− 17

)=(

107

)=(

103

),

and to the equation x1 + x2 + x3 = 11 is given by(r + n− 1

r

)=(

11 + 3− 111

)=(

1311

)=(

132

).

By M.P., the desired number is given by (103

)(132

).

By A.P., the desired number of nonnegative integer solutions to the equation

(x1 + x2 + x3)(y1 + y2 + y3) = 77

is given by (103

)(132

)+(

92

)(143

).


(x1 + x2 + · · ·+ xn)(y1 + y2 + · · ·+ yn) = p,

where n ∈ N and p is a prime.

PB Since p is a prime, then only either x1 + x2 + · · · + xn = p and y1 + y2 + · · · + yn = 1 orx1 + x2 + · · ·+ xn = 1 and y1 + y2 + · · ·+ yn = p. Hence we have:(

p + n− 1p

)(1 + n− 1

1

)+(

1 + n− 11

)(p + n− 1

p

)= 2n

(p + n− 1

p

).

73. There are 5 ways to express “4” as a sum of 2 nonnegative integers in which the order counts:

4 = 4 + 0 = 3 + 1 = 2 + 2 = 1 + 3 = 0 + 4.

Given r, n ∈ N, what is the number of ways to express r as a sum of n nonnegative integers inwhich the order counts?

74. There are 6 ways to express “5” as a sum of 3 positive integers in which the order counts:

5 = 3 + 1 + 1 = 2 + 2 + 1 = 2 + 1 + 2 = 1 + 3 + 1 = 1 + 2 + 2 = 1 + 1 + 3.

Given r, n ∈ N with r ≥ n, what is the number of ways to express r as a sum of n positive integersin which the order counts?

75. A positive integer d is said to be ascending if in its decimal representation: d = dmdm−1 · · · d2d1, wehave

0 < dm ≤ dm−1 ≤ · · · ≤ d2 ≤ d1.

For instance, 1337 and 2455566799 are ascending integers. Find the number of ascending integerswhich are less than 109.

TT For this problem, we can only have ascending integers with at most nine digits. To determineall such integers, let us consider a seqeunce of 9 sticks and 9 “+” signs.

If we were to initially place all nine “+” signs in a row, there will be 10 spaces available for the sticksto occupy (8 spaces in between the “+” signs and 2 spaces at the ends of the row). Relating thisto the given problem, we can actually view the first 9 spaces as the numbers from 1 to 9 (i.e., the

Exercises 1 of 32


first space represents the digit 1, the 2nd space represents the digit 2, and so on). As for the 9sticks, they determine the number of digits that the ascending number will have.

In particular, if we let 7 of the sticks occupy (in any manner) the first 9 spaces — and thereforethe 2 remaining sticks occupy the last space — this means that we’re actually forming a 7-digitnumber. If only 1 stick occupies any of the first 9 spaces — and therefore the 8 remaining stickswill occupy the last space — this means that we’re just forming a 1-digit number. Notice that themore sticks placed in the last space, the lesser the number of digits that the number will have.Moreover, we can only construct a number with at most 9 digits, since we only have nine sticks.

To illustrate this, consider the sequence

||+ |||+ |||+ + + + + + + |.

Using the above interpretation, we can say that the number contains 8 digits, composed of two1s, three 2s, three 3s. Note, however, that there is only one way in which we can construct anascending number, and that is 11222333.

Here’s another sequence:+ + |+ +|+ + + + + |||||||.

This can be interpreted as an ascending number containing only two digits, and it is equal to 35.

Thus, the number of ways of arranging 9 sticks and 9 “+” signs can give us the total number ofascending numbers less than 109, which is

(189

). However, we have to eliminate the sequence

+ + + + + + + + +||||||||||,

since it does not correspond to any ascending number.

Thus, the total number of ascending numbers less than 109 is(189

)− 1.

76. A positive integer d is said to be strictly ascending if in its decimal representation:d = dmdm−1 · · · d2d1, we have

0 < dm < dm−1 < · · · < d2 < d1.

For instance, 145 and 23689 are strictly ascending integers. Find the number of strictly ascendingintegers which are less than (i) 109, (ii) 105.

77. Let A = {1, 2, . . . , n}, where n ∈ N.

(i) Given k ∈ A, show that the number of subsets of A in which k is the maximum number isgiven by 2k−1.

(ii) Apply (i) to show thatn−1∑i=0

2i = 2n − 1.

78. In a given circle, n ≥ 2 arbitrary chords are drawn such that no three are concurrent within theinterior of the circle. Suppose m is the number of points of intersection of the chords within theinterior. Find, in terms of n and m, the number r of line segments obtained through dividing thechords by their points of intersection.

79. There are p ≥ 6 points given on the circumference of a circle, and every two of the points are joinedby a chord.

(i) Find the number of such chords.

Assume that no 3 chords are concurrent within the interior of the circle.

(ii) Find the number of points of intersection of these chords within the interior of the circle.

(iii) Find the number of line segments obtained through dividing the chords by their points ofintersection.

Exercises 1 of 32


(iv) Find the number of triangles whose vertices are the points of intersection of the chords withinthe interior of the circle.

80. In how many ways can n + 1 different prizes be awarded to n students in such a way that eachstudent has at least one prize?

FO Since there are n students, then they can be arranged n! ways for the prizes to be awarded.The fact that at least one prize each student then

(n+1

r

)be number of distributions to be awarded

at n students, where r = 2, for r > 2 there will be student with no prize at all. Thus(n+1

2

)n! by (MP)

are the total number of ways.

81. (a) Let n, m, k ∈ N, and let Nk = {1, 2, . . . , k}. Find

(i) the number of mappings from Nn to Nm.(ii) the number of 1-1 mappings from Nn to Nm, where n ≤ m.

(b) A mapping f : Nn → Nm is strictly increasing if f(a) < f(b) whenever a < b in Nn. Find thenumber of strictly increasing mappings from Nn to Nm, where n ≤ m.

(c) Express the number of mappings from Nn onto Nm in terms of S(n, m) (the Stirling numberof the second kind).

82. Given r, n ∈ Z with 0 ≤ n ≤ r, the Stirling number s(r, n) of the first kind is defined as the numberof ways to arrange r distinct objects around n identical circles such that each circle has at least oneobject. Show that

(i) s(r, 1) = (r − 1)! for r ≥ 1;

(ii) s(r, 2) = (r − 1)!(1 + 12 + 1

3 + · · ·+ 1r−1 ) for r ≥ 2;

(iii) s(r, r − 1) =(r2

)for r ≥ 2;

(iv) s(r, r − 2) = 124r(r − 1)(r − 2)(3r − 1) for r ≥ 2;

(v)∑r

n=0 s(r, n) = r!.

83. The Stirling numbers of the first kind occur as the coefficients of xn in the expansion of

x(x + 1)(x + 2) · · · (x + r − 1).

For instance, when r = 3,

x(x + 1)(x + 2) = 2x + 3x2 + x3 = s(3, 1)x + s(3, 2)x2 + s(3, 3)x3;

and when r = 5,

x(x + 1)(x + 2)(x + 3)(x + 4) = 24x + 50x2 + 35x3 + 10x4 + x5

= s(5, 1)x + s(5, 2)x2 + s(5, 3)x3 + s(5, 4)x4 + s(5, 5)x5.

Show that

x(x + 1)(x + 2) · · · (x + r − 1) =r∑

n=0

s(r, n)xn,

where r ∈ N.

84. Given r, n ∈ Z with 0 ≤ n ≤ r, the Stirling number S(r, n) of the second kind is defined as thenumber of ways of distributing r distinct objects into n identical boxes such that no box is empty.Show that

(i) S(r, 2) = 2r−1 − 1;

(ii) S(r, 3) = 12 (3r−1 + 1)− 2r−1;

(iii) S(r, r − 1) =(r2

);

(iv) S(r, r − 2) =(r3

)+ 3(r4

).

Exercises 1 of 32


85. Let (x)0 = 1 and for n ∈ N, let

(x)n = x(x− 1)(x− 2) · · · (x− n + 1).

The Stirling numbers of the second kind occur as the coefficients of (x)n when xr is expressed interms of (x)ns. For instance, when r = 2, 3 and 4, we have, respectively,

x2 = x + x(x− 1) = (x)1 + (x)2= S(2, 1)(x)1 + S(2, 2)(x)2

x3 = x + 3x(x− 1) + x(x− 1)(x− 2)= S(3, 1)(x)1 + S(3, 2)(x)2 + S(3, 3)(x)3

x4 = x + 7x(x− 1) + 6x(x− 1)(x− 2) + x(x− 1)(x− 2)(x− 3)= S(4, 1)(x)1 + S(4, 2)(x)2 + S(4, 3)(x)3 + S(4, 4)(x)4

Show that for r = 0, 1, 2, . . . ,

xr =r∑

n=0

S(r, n)(x)n.

86. Suppose that m chords of a given circle are drawn in such a way that no three are concurrent inthe interior of the circle. If n denotes the number of points of intersection of the chords within thecircle, show that the number of regions divided by the chords in the circle is m + n + 1.

87. For n ≥ 4, let r(n) denote the number of interior regions of a convex n-gon divided by all itsdiagonals if no three diagonals are concurrent within the n-gon. For instance, as shown in thefollowing diagrams, r(4) = 4 and r(5) = 11. Prove that r(n) =

(n4

)+(n−1

2

).

��@

@@ �

��

LLLLL

,,,,,

��

CCCCCCC

ll

ll

l��

ZZ

ZZ

88. Let n ∈ N. How many solutions are there in ordered positive integer pairs (x, y) to the equation

xy

x + y= n?

PB Let n, x and y be elements of postive integers. Then the equation

xy

x + y= n ⇐⇒ xy = n(x + y) ⇐⇒ xy − nx = ny ⇐⇒ x(y − n) = ny (88a)

Thus, ny > 0 and so as x(y − n) > 0. This implies that y − n > 0, which enables us to write y interms of n, i.e., y = n + b, for some positive integer b.

(88a) ⇐⇒ x =ny

y − n,

substituting n + a for y gives us:

x =n(n + b)

b⇐⇒ x =

n2

b+ n (88b)

However, x is a positive integer, then for (88b) to be valid, then b must divide n2, or in other words,b must be a factor of n2. Thus this answers the problem: the number of solutions to the originalproblem is the same as the number of positive factors or divisors of n2.

Furthermore, let n = p1k1 · p2

k2 · · · pmkm , where pis are distint prime numbers and kis are positive

integers. Then n2 would have (2k1 + 1)(2k2 + 1) · · · (2km + 1) factors. This is also the number ofsolutions to the problem given above.

Exercises 1 of 32


89. Let S = {1, 2, 3, . . . , 1992}. In each of the following cases, find the number of 3-element subsets{a, b, c} of S satisfying the given condition:

(i) 3 | (a + b + c);PB Express S = {1, 2, 3, . . . , 1992} in terms of its remainder when divided by 3, that is, wecan write it in its “modulo-3” form, say S′ = {1, 2, 0, 1, 2, 0, . . . , 1, 2, 0}, where each numberfrom S is represented in S′ by its remainder when divided by 3. Thus we have 664 elementsof 1s, 2s and 0s in S′. We denote the set of all 1s by set A, all 2s by set B, and all 0s by setC. Therefore, to have the sum of any 3 elements from S′ divisible by 3, we can simply chooseall three numbers from the one set, i.e., all from either set A, set B or set C; or we chooseone number from each of the three sets, i.e., one number from set A, one from set B and onefrom set C. Therefore, the number of ways to do this is given by

3(

6643

)+(

6641

)3

.

(ii) 4 | (a + b + c).PB Similarly, let S′′ = {1, 2, 3, 0, 1, 2, 3, 0, . . . , 1, 2, 3, 0}, where each number from S is repre-sented in S′′ by its remainder when divided by 4. Thus we have 498 elements of 1s, 2s, 3sand 0s in S′′. Again, we denote the set of all 1s by set A, all 2s by set B, all 3s by set C, andall 0s by set D. In order to have the sum of any three numbers from S′′, we consider threecases. First, we can choose all three numbers from set D; second, we can choose exactlyone number from set A, set C and set D, respectively; finally, we can choose two numbersfrom set C and one number from set B or two numbers from set A and one number from setB or two numbers from set B and one from set D. Therefore, the number of ways to do thisis given by (

4983

)+(

4981

)3

+ 3(

4981

)(4982

).

90. A sequence of 15 random draws, one at a time with replacement is made from the set

{A, B, C, . . . , X, Y, Z}

of the English alphabet. What is the probability that the string

UNIVERSITY

occurs as a block in the sequence?

PB The number of ways the string UNIVERSITY appears as a block is 265 · 6 and the numberof ways to draw 15 letters with replacement is 2615. Thus the probability is 265 · 6/2615 or simply6/2610.

91. A set S = {a1, a2, . . . , ar} of positive integers, where r ∈ N and a1 < a2 < · · · < ar, is said tobe m-separated (m ∈ N) if ai − ai−1 ≥ m, for each i = 2, 3, . . . , r. Let X = {1, 2, . . . , n}. Find thenumber of r-element subsets of X which are m-separated, where 0 ≤ r ≤ n− (m− 1)(r − 1).

MS It is evident that there is only one 0-element subset of X, ∅, and there are n 1-elementsubsets of X, and all these subsets are vacuously m-separated. Let x1 = a1 and xi = ai − ai−1

for 1 < i < r + 1 and xr+1 = n− ar. Note that x1 ≥ 1, xr+1 ≥ 0, and for 1 < i < r + 1, xi ≥ m andx1 + x2 + · · · + xr + xr+1 = n. There is a bijection between the collection of solutions S and thecollection of solutions x1, x2, . . . , xr+1; there is also a bijection between the collection of solutionsx1, x2, . . . , xr+1 and the collection of solutions y1, y2, . . . , yr+1 such that yi ≥ 0 for 1 ≤ i ≤ r+1 andy1 + · · · + yr+1 = n − 1 − (r − 1)m, which has cardinality Hr+1

n−1−(r−1)m =(n−1−(r−1)m+r+1−1

n−1−(r−1)m

)=(

n−(r−1)(m−1)r

). Note: when r = 0, H1

n+m−1 =(n+m−1

0

)= 1; when r = 1, H2

n−1 =(n1

)= n.

92. Let a1, a2, . . . , an be positive real numbers, and let Sk be the sum of products of a1, a2, . . . , an takenk at a time. Show that

SkSn−k ≥(

n

k

)2

a1a2 · · · an,

Exercises 1 of 32


for k = 1, 2, . . . , n− 1.

TT First, we shall construct two sets S and T in the following manner:

We will choose k positive integers from the set a1, a2, . . . , an, and get the product. Let that be s1,which shall become an element in S. For the remaining n−k numbers not chosen, let their productbe t1. The resulting term shall become an element in T . Note that s1 · t1 = a1a2a3 · · · an.

If we take a different set of k positive integers from a1, a2, . . . , an, we will again have to two products,one of which will become a member of S (say s2) and the other will become a member of T (sayt2). Again, the product s2 · t2 = a1a2a3 · · · an.

The process will continue until we have chosen all possible combinations of k positive integersfrom the set a1, a2, . . . , an, which will allow us to fill in the sets S and T . Note that this processwill be done

(nk

)times (referring to the number of possible ways k number may be chosen from n

positive integers).

Ultimately, we will have sets S = {s1, s2, . . . , s(nk)} and T = {t1, t2, . . . , t(n

k)}, where si · ti =

a1a2a3 · · · an for i = 1, 2, . . . ,(nk

).

Relating this to the given problem, we can actually say that Sn =(n

k)∑i=1

si. The addends that constitute

the sum Sn are always a product of k positive integers taken from a1, a2, . . . , an, which can be saidfor each element in set S.

As for Sn−k, the products of n− k integers that make up its sum can be related to the elements ofset T (where each element is a product of n − k integers). While Sn−k is a sum of

(n

n−k

)terms,

we know that(nk

)is equal to

(n

n−k

). Thus, we can represent Sn−k as the sum of

(nk

)products, all

of which coming from the set T — that is, Sn−k =(n

k)∑i=1

ti.

To prove the inequality, we shall use two methods: the Generalized AM-GM Inequality and theCauchy Schwarz Inequality.

Method I: The Generalized AM-GM Inequality is stated as follows:For nonnegative numbers a1, a2, . . . , an,

1n

n∑i=1

ai ≥ n√

a1a2 · · · an.

Since all elements of S and T are positive, then we can establish the following statementsbased on the Generalized AM-GM inequality,

Sn(nk

) ≥ (s1s2 · · · s(nk))1/(n

k) Sn−k(nk

) ≥(t1t2 · · · t(n

k))1/(n

k),

which can be converted to

Sn ≥(

n

k

)(s1s2 · · · s(n

k))1/(n

k)Sn−k ≥

(n

k

)(t1t2 · · · t(n

k))1/(n

k).

Combining the two inequalites, we shall have

SnSn−k ≥(

n

k

)2((s1 · t1)(s2 · t2) · · · (s(n

k) · t(nk)))1/(n

k)=(

n

k

)2((a1a2a3 · · · an)(

nk))1/(n

k),

since si · ti = a1a2a3 · · · an for i = 1, 2, . . . ,(nk

); then

SnSn−k ≥(

n

k

)2

a1a2a3 · · · an,

which is the desired inequality.

Exercises 1 of 32


Method II: The Cauchy-Schwarz Inequality is stated as follows:For real numbers x1, x2, . . . , xn, y1, y2, . . . , yn, we have(

n∑i=1

xiyi

)2

≤

(n∑

i=1

xi2

)(n∑

i=1

yi2

).

Since si, ti ≥ 0, then√

si and√

ti are real numbers for i = 1, 2, . . . ,(nk

). Hence, we can use

CS Inequality: (nk)∑

i=1

√si

√ti

2

≤

(nk)∑

i=1

si

(n

k)∑i=1

ti

.

Note that si · ti = a1a2a3 · · · an for i = 1, 2, . . . ,(nk

).(n

k)∑i=1

√siti

2

≤

(nk)∑

i=1

si

(n

k)∑i=1

ti

(n

k)∑i=1

√a1a2a3 · · · an

2

≤

(nk)∑

i=1

si

(n

k)∑i=1

ti

((

n

k

)√

a1a2a3 · · · an

)2

≤

(nk)∑

i=1

si

(n

k)∑i=1

ti

(

n

k

)2

a1a2a3 · · · an ≤ SnSn−k,

which is the desired inequality.

93. For {1, 2, 3, . . . , n} and each of its nonempty subsets, a unique alternating sum is defined as follows:Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alter-nately add and subtract successive numbers. (For example, the alternating sum for {1, 2, 4, 6, 9} is9−6+4−2+1 = 6 and for {5} it is simply 5.) Find the sum of all such alternating sums for n = 7.

MS Let n ∈ N, A ⊆ Nn = {1, 2, . . . , n}, aA be the alternating sum of A, a∅ = 0, P(X) denote thepower set of X, Nn = {A ∪ {n} : A ∈ P(Nn−1)} = P(Nn) \ P(Nn−1) and

An =∑

A∈P(Nn)

aA — claim that An = n2n−1.

For each A ∈ P(Nn−1), there is a unique B = A ∪ {n} ∈ Nn; for each B ∈ Nn, there is a uniqueA = B \{n} ∈ P(Nn−1). If A ∈ P(Nn−1), then aA∪{n} = n−aA, since n is larger than the elementsof A, and those elements will change signs relative to their sign in aA. Thus, since |P(Nk)| = 2k,

An =∑

A∈P(Nn)

aA =∑

A∈P(Nn−1)

aA +∑

B∈Nn

aB =∑

A∈P(Nn−1)

(aA + aA∪{n}) =∑

A∈P(Nn−1)

n = n2n−1.

Therefore, the sum An of all such alternating sums for n = 7 is A7 = 7 · 27−1 = 448.

94. A gardener plants three maple trees, four oak trees and five birch trees in a row. He plants themin random order, each arrangement being equally likely. Let m

n in lowest terms be the probabilitythat no two birch trees are next to one another. Find m + n.

95. In a tournament, each player played exactly one game against each of the other players. In thegame the winner was awarded 1 point, the loser got 0 points, and each of the two players earned1/2 point if the game was a tie. After the completion of the tournament, it was found that exactly

Exercises 1 of 32


half of the points earned by each player were earned in games against the ten players with theleast number of points. (In particular, each of the ten lowest scoring players earned half of her/hispoints against the other nine of the ten.) What was the total number of players in the tournament?

MS Let there be be n players in the tournament. Of the lowest ten players,(102

)= 45 matches

were played, and thus, the lowest ten players scored a total of 45 points amongst themselves and45 points with the other n − 10 players. There were

(n−10

2

)matches played amongst the other

players, for as many total points, and(n−10

2

)+45 consititutes half the total points of the n people in

the tournament, since the other half of the points were contributed by matches between people inthe lowest ten and the others. There are

(n2

)total matches, and thus the same number of points.

Thus, (n

2

)=

n(n− 1)2

= 2((

n− 102

)+ 45

)= (n− 10)(n− 11) + 90

n2 − n = 2n2 − 42n + 400n2 − 41n + 400 = 0 =⇒ n = 16 or n = 25.

If n = 16, then the top six players would only have an average of 2(62

)/6 = 5 points, which would

be less than the average score of the lowest ten players, which is 2 · 45/10 = 9 points. Thus, therewere 25 players in the tournament.

96. Let S be the sum of the base 10 logarithms of all of the proper divisors of 1,000,000. (By a properdivisor of a natural number we mean a positive integral divisor other than 1 and the number itself.)What is the integer nearest to S?

TT The prime factorization of 1,000,000 is given by 26·56; based on this, we can see that 1,000,000has a total of 7 · 7 = 49 positive divisors. Let the divisors be denoted by (in no particular order ofmagnitude) d1, d2, . . . , d49.

Since S is the sum of the base 10 logarithms of all the proper divisors of 1,000,000 (which excludes1 and 1,000,000), then S can be expressed as

S = log d1 + log d2 + · · ·+ log d49 − log 1− log 1000000 = logd1d2 · · · d49

1000000.

To determine S, we should find the product d1d2 · · · d49.

Note that each of the 49 divisors of 1,000,000 can be expressed in the form 2x ·5y, for 0 ≤ x, y ≤ 6.This means that the product d1d2 · · · d49 can then be expressed in the form 2m ·5n, where m,n ∈ N.

To solve for m, we consider that: (a) 7 of the divisors have the form 21 · 5y (21 · 50, 21 · 51, . . . ,21 · 56); (b) 7 of the divisors have the form 22 · 5y; (c) 7 of the divisors have the form 23 · 5y; (d) 7 ofthe divisors have the form 24 · 5y; (e) 7 of the divisors have the form 25 · 5y; and (f) 7 of the divisorshave the form 26 · 5y. Considering all these cases, we can say that

2m = (21)7 · (22)7 · (23)7 · (24)7 · (25)7 · (26)7 = 2147,

which means m = 147. To solve for n, we may employ the same method, and by symmetry, wecan show that

5n = (51)7 · (52)7 · (53)7 · (54)7 · (55)7 · (56)7 = 5147,

which means n = 147. Thus, the product of the divisors d1d2 · · · d49 is equal to 2147 · 5147 = 10147.

Solving for the value of S, we have

S = logd1d2 · · · d49

1000000= log

10147

106= log 10141 = 141.

97. In a sequence of coin tosses one can keep a record of the number of instances when a tail is im-mediately followed by a head, a head is immediately followed by a head, etc. We denote theseby TH , HH , etc. For example, in the sequence HHTTHHHHTHHTTTT of 15 coin tosses weobserve that there are five HH , three HT , two TH and four TT subsequences. How many dif-ferent sequencesof 15 coin tosses will contain exactly two HH , three HT , four TH and five TTsubsequences?

Exercises 1 of 32


98. An ordered pair (m,n) of nonnegative integers is called simple if the addition m + n in base 10requires no carrying, Find the number of simple ordered pairs of nonnegative integers that sum to

(i) 1492;PB Let abcd and efgh be two “simple” 4-digit numbers (the first digit can be zero, in that case,the number becomes 3-digit only). We want to find the number of “simple” integers that sumup to 1492. To do this, let abcd + efgh = 1492. Since abcd and efgh are “simple” integers,then no carrying is allowed in addition. Hence, the problem reduces to finding the number ofnon-negative integers to the following linear equations:

a + e = 1 (98a)b + f = 4 (98b)c + g = 9 (98c)d + h = 2 (98d)

It is easy to show that (98a) has 2 nonnegative integer solutions, (98b) has 5, (98c) has 10and (98d) has 3. Thus, by multiplication property, there are a total of 2 · 5 · 10 · 3 = 300 numberof simple nonnegative integer solutions to abcd + efgh = 1492.

(ii) 1992.PB Similarly, the problem reduces to finding the number of non-negative integers to thefollowing linear equations:

a + e = 1 (98a)b + f = 4 (98b)c + g = 9 (98c)d + h = 2 (98d)

(98a) has 2 non-negative integer solutions, (98b) has 10, (98c) has 10 and (98d) has 3. Thus,by multiplication property, there are a total of 2 ·10 ·10 ·3 = 600 number of simple nonnegativeinteger solutions to abcd + efgh = 1992.

99. Let m/n, in lowest terms, be the probability that a randomly chosen positive integer of 1099 is aninteger multiple of 1088. Find m + n.

PB 1099 can be expressed as 299 · 599, thus 1099 has 1002 positive divisors. Integer multiples of1088 can be expressed in the form 2x · 5y, where x and y are both integers and 88 ≤ x, y ≤ 99.Thus among the 10000 positive divisors of 1099, 12 ·12 are integer multiples of 1088. Therefore, theprobability that a randomly chosen positive divisor of 1099 is an integer multiple of 1088 is 144/10000or 9/625. Thus, m + n = 634.

100. A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons.At each vertex of the polyhedron, one square, one hexagon, and one octagon meet. How manysegments joining vertices of the polyhedron lie in the interior of the polyhedron rather than alongan edge or a face?

MS It is evident that no two polygons of the same shape share a side, and thus there are 48vertices on the polygon. Each vertex shares a face with 3 + 5 + 7 − 3 = 12 other vertices, sinceeach of the faces that the vertex is on must share a side with an adjacent face. Thus, there are48 − 13 = 35 other vertices that do not share a face with that vertex, and the edge connectingany of these vertices to the given point must lie in the interior of the polyhedron. Since eachedge is counted twice (once per vertex it is incident on), the number of required edges must be48 · 35/2 = 840.

101. Someone observed that 6! = 8 · 9 · 10. Find the largest positive integer n for which n! can beexpressed as the product of n− 3 consecutive positive integers.

MS It is apparent that the largest number in the n − 3 consecutive positive integers multiplied is

Exercises 1 of 32


larger than n, and larger than 3. Let n+r be the largest number among the positive integers. Thenthe problem becomes solving for n in

(n + r)(n + r − 1) · · · (n + r − (n− 4)) =(n + r)!(r + 3)!

= n! =⇒ n!3!

r∏i=1

n + r

3 + r= n! =⇒

r∏i=1

n + r

3 + r= 6.

Since n > 3, the fraction (n + r)/(3 + r) > 1, and the larger r becomes, the smaller (n + 1)/4becomes, and the smaller n becomes. Thus, n is largest when r = 1, and n + 1 = 24, n = 23.

102. Let S = {1, 2, . . . , n}. Find the number of subsets A of S satisfying the following conditions:

A = {a, a + d, . . . , a + kd} for some positive integers a, d and k, andA ∪ {x} is no longer an arithmetic progression with common difference d for each x ∈ S \A.

103. Find all natural numbers n > 1 and m > 1 such that

1! · 3! · 5! · · · (2n− 1)! = m!.

104. Show that for n ∈ N,n∑

r=0

Pnr = bn!ec,

where bxc denotes the greatest integer ≤ x and e = 2.718 . . . .

PB First verify for n = 1:

1∑r=0

Pnr =

1!(1− 0)!

+1!

(1− 1)!= 2 = b1!(e)c.

For n ≥ 2, consider the power (Maclaurin) series of ex,

ex = 1 + x +x2

2!+

x3

3!+ · · ·+ xk

k!+ · · ·+ xn

n!+

xn+1

(n + 1)!eξx︸︷︷︸

error function

, where 0 < ξx < x.

Letting x = 1 and multiplying everything by n!,

n!e = n! + n! +n!2!

+ · · ·+ n!k!

+ · · ·+ n!(n− 1)!

+n!n!

+n!

(n + 1)!eξ1

= Pnn + Pn

n−1 + Pnn−2 + · · ·+ Pn

n−k + · · ·+ Pn1 + Pn

0 +n!

(n + 1)!eξ1 .

However,

n!(n + 1)!

eξ1 =1

n + 1eξ1 <

1n + 1

e since 0 < ξ1 < 1

< 1 since e < n + 1 for all n ≥ 2.

Thus,

bn!ec =⌊Pn

n + Pnn−1 + · · ·+ Pn

1 + Pn0 +

n!(n + 1)!

eξ1

⌋= Pn

n + Pnn−1 + · · ·+ Pn

1 + Pn0 =

n∑r=0

Pnr .

105. Let S = {1, 2, . . . , 1990}. A 31-element subset A of S is said to be good if the sum∑

a∈A a is divisibleby 5. Find the number of 31-element subsets of S which are good.

106. Let S be a 1990-element set and let P be a set of 100-ary sequences (a1, a2, . . . , a100), where ais aredistinct elements of S. An ordered pair (x, y) of elements of S is said to appear in (a1, a2, . . . a100)if x = ai and x = aj for some i, j with 1 ≤ i < j ≤ 100. Assume that every ordered pair (x, y) ofelements of S appears in at most one member in P . Show that

|P| ≤ 800.

Exercises 1 of 32


107. Let M = {r1 ·a1, r2 ·a2, . . . , rm ·an} be a multiset with r1 + r2 + · · ·+ rn = r. Show that the numberof r-permutations of M is equal to the number of (r − 1)-permutations of M .

TT The number of r-permutations of M is given byr!

r1!r2! · · · rm!. As for the number of r − 1-

permutations of M , we shall consider the r− 1-permutations of several multisets of the form Mi ={r1 · a1, r2 · a2, . . . , (ri − 1) · ai, . . . , rm · am} for i = 1, 2, . . . ,m, where |Mi| = r − 1.

Counting the total number of permutations for all Mi,

m∑i=1

(r − 1)!r1!r2! · · · (ri − 1)! · · · rm!

= S.

Manipulating the expression,

S =(r − 1)!

(r1 − 1)!r2! · · · rm!+

(r − 1)!r1!(r2 − 1)! · · · rm!

+ · · ·+ (r − 1)!r1!r2! · · · (rm − 1)!

=(r − 1)!

(r1 − 1)!(r2 − 1)! · · · (rm − 1)!

(1

r2r3 · · · rm+

1r1r3 · · · rm

+ · · ·+ 1r1r2 · · · rm−1

)=

(r − 1)!(r1 − 1)!(r2 − 1)! · · · (rm − 1)!

(r1 + r2 + · · ·+ rm

r1r2r3 · · · rm

).

Sincem∑

i=1

ri = r,

S =(r − 1)!

(r1 − 1)!(r2 − 1)! · · · (rm − 1)!

(r

r1r2r3 · · · rm

)=

r!r1!r2! . . . rm!

.

Thus the number of r-permutations of M is equal to the number of r − 1-permutations of M .

108. Prove that it is impossible for seven distinct straight lines to be situated in the Euclidean plane soas to have at least six points where exactly three of these lines intersect and at least four pointswhere exactly two of these lines intersect.

109. For what n ∈ N does there exist a permutation (x1, x2, . . . , xn) of (1, 2, . . . , n) such that the differ-ences |xk − k|, 1 ≤ k ≤ n, are all distinct?

110. Numbers d(n, m), where n, m are integers and 0 ≤ m ≤ n, are defined by

d(n, 0) = d(n, n) = 1 for all n ≥ 0

andm · d(n, m) = m · d(n− 1,m) + (2n−m) · d(n− 1,m− 1)

for 0 < m < n. Prove that all the d(n, m) are integers.

111. A difficult mathematical competition consisted of a Part I and a Part II with a combined total of28 problems. Each contestant solved 7 problems altogether. For each pair of problems, there wereexactly two contestants who solved both of them. Prove that there was a contestant who, in Part I,solved either no problems or at least four problems.

112. Suppose that five points in a plane are situated so that no two of the straight lines joining themare parallel, perpendicular, or coincident. From each point perpendiculars are drawn to all thelines joining the other four points. Determine the maximum number of intersections that theseperpendiculars can have.

113. Let n distinct points in the plane be given. Prove that fewer than 2n32 pairs of them are at unit

distance apart.

114. If c and m are positive integers each greater than 1, find the number n(c,m) of ordered c-tuples(n1, n2, . . . , nc) with entries from the initial segment {1, 2, . . . ,m} of the positive integers such thatn2 < n1 and n2 ≤ n3 ≤ · · · ≤ nc.

Exercises 1 of 32


115. Let X = {x1, x2, . . . , xm}, Y = {y1, y2, . . . , yn}, (m,n ∈ N) and A ⊆ X × Y . For xi ∈ X , let

A(xi, ·) = ({xi} × Y ) ∩A

and for yi ∈ Y , letA(·, yi) = (X × {yi}) ∩A.

(i) Prove the following Fubini Principle:

m∑i=1

|A(xi, ·)| = |A| =n∑

j=1

|A(·, yj)|.

(ii) Using (i), or otherwise, solve the following problem: There are n ≥ 3 given points in the planesuch that any three of them form a right-angled triangle. Find the largest possible value of n.

Source: Chen, Chuan-Chong and Koh, Khee-Meng. Principles and Techniques in Combinatorics.Reproduced without explicit permission, for use in Math 248 class, Ateneo de Manila University, Philippines, 1st Semester AY2006-2007.

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