Permutations: Practise Problems

1. How many ways can you arrange the letters in the word REDCOATS...

a) with no restrictions

b) if the word must start with a vowel?

c) if it must start with a consonant?

d) if the 2nd, 5th, and 7th letters must be vowels?

e) if it must start with a vowel and end with a consonant

f) if the smaller word RED must appear in the arrangement

g) if all of the vowels must not be together

2. How many 8 letter words can be made from the word ABSOLUTE...

a) with no restrictions

b) if it must start with an A

c) if A and B must be together

d) if it A and B must be together at the front of the word

e) if B must always follow A

f) if ABS must be together

3. In the annual Factorial Road Race how many ways can...

a) the cars finish if there are 10 racers?

b) 1st

, 2nd

, and 3rd

be determined if there are 10 racers?

c) 1st

, 2nd

, and 3rd

be determined if there are 20 Racers?

d) the top 20 cars finish if there are 30 racers?

4. How many ways can 5 people be seated on a bench?

5. How many ways can 3 girls and three boys sit in a row if boys and girls must alternate?

6. How many arrangements are there of...

a) 30 students in a classroom with 30 desks

b) 17 students in a classroom with 25 desks

c) 4 ski tows up the hill and 7 routes down the hill

d) a room if the interior decorator has selected 18 carpet possible carpets, 25 wall patterns, and 7 paint colours

7. Twenty students are available for 7 scholarships. How many ways can the scholarships be awarded if...

a) a student can only be granted one scholarship

b) a student can be granted more than one scholarship

8. Emilio has picked up his textbooks for the seven courses he will study this year. In how many ways can he arrange them

on his bookshelf if he wants to keep the French and the German texts side by side?

9. How many 4 digit numbers can be made using 0 - 7 with no repeated digits allowed?

10. How many 3-digit numbers are there in which the tens and the ones digits are different?

11. As an anniversary gift a family with 4 children give their parents a photo montage containing 5 photos, including one

family portrait and 1 of each of the children.

a) In how many ways could the photos be hung in a row on the living room wall?

b) If the family portrait must be in the middle, how many orders can the pictures be hung on the wall?

12. How many 4 digit numbers, using the numbers 1 - 6, no more than once each,

a) do not have a 3?

b) have a 1 on the third position

c) have a 3 in the second position and a 5 in the third position

13. The Canadian postal code system consists of six characters where the first, third, and fifth characters are letters and the

other characters are digits.

a) No postal code in Canada can begin with the letters D,F,I,O,Q,U, but repeated letters are allowed and any digit

is allowed. How many postal codes are possible in Canada?

b) The first letter of the postal code represents the province. In Ontario all postal codes start with K, L, M, N or P.

How many different postal codes are possible in Ontario?

c) How many different postal codes are possible in the rest of Canada?

d) The first digit (or second character) is a 0 if the region it represents is rural. How many different rural postal

codes are possible in Ontario?

e) According to Statistics Canada, only 12% of the available postal codes are in use in 2007. How many postal

codes were in use in 2007 in Canada?

14. A track and field event has 4 time slots, each offering the same 12 sporting events. If students must choose 3 different

event for their day, how many different days can a student plan?

15. How many 5 digit odd numbers can be made from the digits 1 to 7...

a) with repeated numbers allowed

b) with no repeated numbers allowed

16. Using 0-9, how many

a) 4 digit numbers are evenly divisible by 5 with repeated digits allowed?

b) 4 or 5 digit numbers are evenly divisible by 5 with repeated digits allowed?

c) 6 digit numbers are evenly divisible by 5 with repeated digits not allowed?

17. There are 10 teams in the Eastern Division of the Ontario Hockey league.

a) How many different finishes could there be at the end of the season?

b) How many ways can the season end if we know the Niagara Ice Dogs will finish first, the Sudbury Wolves will

finish 7th

and the Mississauga St. Mike’ Majors finish last?

c) How many finishes could there be if we know that Niagara, Sudbury, or Mississauga will win the Eastern

division?

d) How many ways can the season end if we know that Niagara, Sudbury, and Mississauga will be together in the

standings?

e) How many ways can the season end if we know that Niagara, Sudbury, and Mississauga will be the top 3 teams.

f) How many different finishes can there be if we know Niagara will finish in the top 3?

18. Laura lost Jordan’s phone number. All she remembers is that it didn't have a 0 or 1 in the first three digits.

a) How many seven-digit telephone numbers are possible if repeated digits are allowed?

b) Laura remembers that not only was there no 0's or 1's in the first three digits, but there were no repeated

numbers in the phone number. How many less combinations will she have to try now?

19. 5 Boys and 8 Girls go to the Halloween dance. Lori and Julie will not dance with Bert or Tyler. Charles will not dance with

Diana or Sara. How many boy/girl couples can be made?

20. Lotto Max draws 7 numbers from a 1 to 49.

a) How many ways can the 7 numbers be drawn, in order?

b) Since order doesn't matter, how many different combinations of winning numbers can there be?

Permutations: Practise Problems (Solutions)

1. a) 8! = 40,320

b) 3 vowels: 3 x 7! = 15,120

c) 5 consonants: 5 x 7! = 25,200

d) CVCCVCVC: 5 x 3 x 4 x 3 x 2 x 2 x 1 x 1 = 5! x 3! = 720

e) V _ _ _ _ _ _ C: 3 x 6! x 5 = 10,800

f) Treat RED as 1 letter, making it a 6 letter word: 6! = 720

g) Vowels not together = total - all vowels together

Treat the 3 vowels together as 1 letter: 6! = 720

Ways to arrange the 3 vowels: 3! = 6

Answer: 8! - (6! x 3!) = 40,320 - (4320) = 36,000

2. How many 8 letter words can be made from the word ABSOLUTE

a) 8! = 40,320

b) 1 x 7! = 5040

c) treat A and B as one letter

2 x 7! = 10,080 (x 2 because of AB or BA)

d) 2! x 6! = 1440

e) treat AB as one letter: 7! = 5040

f) treat ABS as one letter: 6! = 720

ways to arrange ABS: 3! = 6

Answer: 6! x 3! = 4320

3. a) 10! = 3,628,800

b) P(10,3) = 720

c) P(20,3) = 6840

d) P(30,20) = 7.31 x 1025

4. 5! = 120

5. Ways to arrange the 3 boys: 3! = 6

Ways to arrange the 3 boys: 3! = 6

Start with boy or girl (2 ways)

Answer: 2 x 3! x 3! = 72

6. a) 30! = 2.65 x 1032

b) Think of it as 25 students, where 8 of them are identical (representing the empty desks)

25!/8! = 3.85 x 1020

You could also consider this as 25 choices of desk for the 1st student, 24 for the 2nd, etc. P(25,17).

c) 4 x 7 = 28 (straight product rule)

d) 18 x 25 x 7 = 3150 (product rule)

7. a) P(20,7) = 390,700,800

b) 20 x 20 x 20 x 20 x 20 x 20 x 20 = 207 = 1,280,000,000

8. Treat German and French as 1 item: 6! = 720

Can either be GF or FG: 2 choices

Answer: 2 x 6! = 1440

9. Note: The first digit can't be a zero, or it would really be a 3 digit number.

7 choices for first digit, then 7 left for the 2nd, 6 for the 3rd, 5 for the 4th.

7 x 7 x 6 x 5 = 1470 OR 7 x P(7,3) = 1470

10. Total 3 digit numbers (can't start with zero): 9 x 10 x 10 = 900

Total with tens and ones the same: 9 x 10 x 1 = 90 (no choice for ones... it has to be the same as the tens)

Indirect reasoning: 900 - 90 = 810

11. a) 5! = 120

b) 4 x 3 x 1 x 2 x 1 = 4! = 24

12. a) only using 1,2,4,5,6 for the digits: 5 x 4 x 3 x 2 = 5! = 120

b) 5 x 4 x 1 x 3 = P(5,3) = 60 (we start at 5 since the 1 is taken)

c) 4 x 1 x 1 x 3 = 12 (start at 4 since 3 and 5 are taken)

13. a) 20 x 10 x 26 x 10 x 26 x 10 = 13,520,000

b) 5 x 10 x 26 x 10 x 26 x 10 = 3,380,000

c) 13,520,000 - 3,380,000 = 10,140,000

d) 5 x 1 x 26 x 10 x 26 x 10 = 338,000

e) 12% x 13,520,000 = 1,622,400

14. P(12, 3) x 4 = 1320 x 4 = 5280 The P(12, 3) is the number of ways that they can pick the 3 events they will

participate in (in order), and the 4 is to account for the time slot that they'll

take as their break.

15. a) 7 x 7 x 7 x 7 x 4 = 74 x 4 = 9,604 (4 choices for the final number since it must be odd)

b) 6 x 5 x 4 x 3 x 4 = 4 x P(6,4) = 1440 (4 choices for last #, other 4 digits with 6 #s left over)

16. a) First digit can't be zero; last digit must be 0 or 5

9 x 10 x 10 x 2 = 1,800

b) 4 digit #: 1,800 (part a)

5 digit #, same rules as part a: 9 x 10 x 10 x 10 x 2 = 18,000

Total: 1,800 + 18,000 = 19,800

c) Case 1 - ends in 0: 9 x 8 x 7 x 6 x 5 x 1 = P(9,5) = 15,120 (first number can be anything but zero)

Case 2 - ends in 5: 8 x 8 x 7 x 6 x 5 x 1 = 8 x P(8,4) = 13,440 (first number can't be 0 or 5)

Total numbers: 15,120 + 13,440 = 28,560

17. a) 10! = 3,628,800

b) 1 x 7 x 6 x 5 x 4 x 3 x 1 x 2 x 1 x 1 = 7! = 5,040

c) 3 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3 x 9! = 1,088,640 (3 choices for first place)

d) Treat the 3 teams as 1 unit, then the arrangements are: 8! = 40,320

Ways to order the 3 teams: 3! = 6

Answer: 3! x 8! = 541,920

e) 3 x 2 x 1 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3! x 7! = 30,240

f) Case 1 (Ice Dogs 1st): 1 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 9! = 362,880

Case 2 (Ice Dogs 2nd): 9 x 1 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 9! = 362,880

Case 3 (Ice Dogs 3rd): 9 x 8 x 1 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 9! = 362,880

Answer: 3 x 9! = 1,088,640

18. a) 8 x 8 x 8 x 10 x 10 x 10 x 10 = 83 x 104 = 5,120,000

b) 8 x 7 x 6 x 7 x 6 x 5 x 4 = P(8,3) x P(7,4) = 336 x 840 = 282,240 possible numbers

Number of combinations that she eliminated: 5,120,000 - 282,240 = 4,837,760

19. Total possible couples: 5 x 8 = 40

Couple that won't work: 6 (LB, LT, JB, JT, CD, CS)

Answer: 40 - 6 = 34

20. a) P(49,7) = 49!/42! = 4.33 x 1011 (exact # too large for calculator)

b) Number of ways to order the 7 #s: 7!

# of ways to choose 7 numbers (so that order doesn't matter):

Note: Selecting subsets of groups when order doesn't matter (like in part b) is the topic of the next unit

on combinations.