LESSON 8COUNTING

Reminder: Before beginning this lesson remember to redo theproblems from Lesson 4 that you have marked off. Do notunmark a question unless you get it correct.

Strategy 21

Writing a list

Sometimes the easiest way to count the number of possibilities is tosimply list them all. When doing this it is important that we have asystematic way of forming our list. This will reduce the likelihood ofmissing something, or listing something twice.Try to answer the following question by writing a list. Do not check thesolution until you have attempted this question yourself.

LEVEL 5: COUNTING

1.How many integers between 3000 and 4000 have digits thatare all different and that increase from left to right? Solution* Lets form a list:345678967897898935678978989678989789There are 20 integers in this list.

Remarks: (1) Notice that we only wrote down the necessary informationwhen forming our list. For example, the second entry was just written7 instead of 3457. This will save a substantial amount of time.(2) A clear and definite pattern was used in forming this list. In this casethe list was written in increasing order. This will minimize the risk ofduplicating or leaving out entries.

Counting Principle

The counting principle says that if one event is followed by a secondindependent event, the number of possibilities is multiplied.

More generally, if E1, E2 ,...En are n independent events withm1 , m2, ,mn possibilities, respectively, then event E1 followed byevent E2, followed by event E3 followed by event En hasm1 m2 mn possibilities.

Try to answer the following question by using the counting principle. Donot check the solution until you have attempted this question yourself.

LEVEL 3: COUNTING

2.How many integers between 9 and 300 have the tens digitequal to 2, 3, or 4 and the units digit (ones digit) equal to 5 or6? Solution

* There are 2 possibilities for the ones digit (5 or 6). There are 3possibilities for the tens digit (2, 3, or 4). There are 3 possibilities for thehundreds digit (0, 1, or 2).

The counting principle says that we multiply the possibilities to get(2)(3)(3) = 18.

Before we go on try to also solve this problem by writing a list. SolutionLets try to list the numbers in increasing order.

25 125225

26126226

35135235

36136236

45145245

46146246

And thats it. We see that the answer is 18.

Permutations and Combinations

The factorial of a positive integer n, written n! Is the product of allpositive integers less than or equal to n.

n! = 1 *2*3...n0! is defined to be 1, so that n!, is defined for all nonnegative integers n.

A permutation is just an arrangement of elements from a set. Thenumber of permutations of n things taken r at a time is nPr =

For example, the number of permutations of {1, 2, 3} taken 2 at a time is These permutations are 12, 21, 13, 31, 23, and 32.

Note that on the SAT you do not need to know the permutation formula.You can do this computation very quickly on your graphing calculator. Tocompute 3 P 2 , type 3 into your calculator, then in the Math menu scrollover to Prb and select nPr (or press 2). Then type 2 and press Enter. Youwill get an answer of 6.

A combination is just a subset containing a specific number of theelements of a particular set. The number of combinations of n thingstaken r at a time is For example, the number ofcombinations of {1, 2, 3} taken 2 at a time is . Thesecombinations are 12, 13, and 23.

Note that on the SAT you do not need to know the combination formula.You can do this computation very quickly on your graphing calculator. Tocompute , type 3 into your calculator, then in the Math menu scrollover to Prb and select nCr (or press 3). Then type 2 and press Enter. Youwill get an answer of 3.

Note that 12 and 21 are different permutations, but the samecombination.

Example: Compute the number of permutations and combinations ofelements from {a, b, c, d} taken (a) 2 at a time, and (b) 4 at a time.4 P 2= 4!/2! = 12, 4 C 2 = 4!/(2!2!) = 6, 4 P 4 = 4!/0! = 24, 4 C 4 = 4!/(4!0!) = 1

Notes:

(1) The permutations taken 2 at a time are ab, ba, ac, ca, ad, da,bc, cb, bd, db, cd, and dc.

(2) The combinations taken 2 at a time are ab, ac, ad, bc, bd, and cd.

Now see if you can list all 24 permutations of {a, b, c, d} taken 4 at atime. Note that all 24 of these permutations represent the samecombination.

Example: How many committees of 4 people can be formed from agroup of 9?

The order in which we choose the 4 people does not matter. Thereforethis is the combination 9 C 4 = 126.

LEVEL 1: COUNTING

3.A menu lists 7 meals and 5 drinks. How many different meal-drink combinations are possible from this menu?

LEVEL 2: COUNTING

4.Five different books are to be stacked in a pile. In how manydifferent orders can the books be placed on the stack

LEVEL 3: COUNTING

5.A chemist is testing 9 different liquids. For each test, thechemist chooses 4 of the liquids and mixes them together.What is the least number of tests that must be done so thatevery possible combination of liquids is tested?

6.Nine different books are to be stacked in a pile. One book ischosen for the bottom of the pile and another book is chosenfor the top of the pile. In how many different orders can theremaining books be placed on the stack?

LEVEL 4: COUNTING

7. Any 2 points determine a line. If there are 18 points in a plane,no 3 of which lie on the same line, how many lines aredetermined by pairs of these 18 points?

8. Segments , , , and intersect at the labeled pointsas shown in the figure above. Define two points as dependentif they lie on the same segment in the figure. Of the labeledpoints in the figure, how many pairs of dependent points arethere?(A)none(B)3(C)6(D)9(E)12

9.A wall is to be painted one color with a stripe of a differentcolor running through the middle. If 9 different colors areavailable, how many color combinations are possible?

L EVEL 5: C OUNTING

10.A six digit number is to be formed using each of the digits 1, 2,3, 4, 5 and 6 exactly once. How many such numbers are there inwhich the digits 2 and 3 are next to each other?

11. If seven cards, each of a different color are placed in a row sothat the green one is placed at an end, how many differentarrangements are possible?

12. How many positive integers less than 4,000 are multiples of 5and are equal to 13 times an even integer?Answers1. 202. 183. 354. 1205. 1266. 50407. 1538. E9. 7210. 24011. 144012. 30

Full Solutions

5.* Solution using combinations: We are counting the number of ways tochoose 4 of the 9 liquids. This is 9 C 4 = 126.

6.* There are seven books left to stack. Therefore we see that there are7! = (7)(6)(5)(4)(3)(2)(1) = 5040 ways to stack these books.Calculator remark: You can compute 7! in your calculator as follows.After typing 7, simply press MATH, scroll to PRB and then select ! (orpress 4)

7.* Solution using combinations: We need to count the number of waysto choose 2 points from 18. This is the combination 18 C 2 = 153.

8.* Solution using strategy 21: Lets list the dependent pairs of points.

A,F A,B F,D B,C A,D A,CE,F E,D F,B D,C E,B E,C

So there are twelve pairs of dependent points, choice (E).

Note: Notice that our list follows a definite pattern. Here we took onelong segment at a time, and listed first the two pairs of points adjoiningthe two shorter segments, and then the pair adjoining the long segment.

* Solution using combinations: We can count the pairs without actuallymaking a list. There are 4 line segments, each with 3 points. So eachsegment has 3 C 2 = 3 pairs of dependent points. So there are 43 = 12pairs of dependent points all together, choice (E).

9.* Solution using the counting principle: There are 9 ways to choose acolor for the wall. Once this color is chosen there are now 8 ways tochoose a color for the stripe. Therefore there are (9)(8) = 72 possibilities.

Solution using permutations: There are 9 P 2 = 72 ways to choose 2 colorsfrom 9, and place them in a specific order.Important note: Dont let the word combinations in the problem itselftrick you. This is not a combination in the mathematical sense. If youpaint the wall red and the stripe blue, then this is a different choice frompainting the wall blue and the stripe red.

10.* Solution using the counting principle: Lets start by counting thenumber of ways we can place 2 and 3 with 2 to the left of 3. Well, 2 cango in the first, second, third, fourth, or fifth position. So there are 5ways.

Now, there are 5 additional ways we can place 2 and 3 with 2 to theright of 3. So all together there are 5 + 5 = 10 ways to place 2 and 3 sothat these two numbers are next to each other. Once we place 2 and 3,there are four places left. So now there are 4 ways to place 1, 3 ways toplace 4, 2 ways to place 5, and 1 way to place 6.Using the counting principle, we see that the final answer is(10)(4)(3)(2)(1) = 240.

11.* Solution using the counting principle: There are 2 ways to place thegreen card. Once the green card is placed, there are 6 ways to place thenext card, 5 ways to place the card after that, then 4, then 3, then 2, andfinally 1 way to place the last card. By the counting principle there are(2)(6)(5)(4)(3)(2)(1) = 1440 different arrangements.

12.* Note that 13 times an even integer is just a multiple of (13)(2) = 26. Sowe are looking for positive integers less than 4,000 that are multiples ofboth 5 and 26. Since 5 and 26 have no prime factors in common, we arejust looking for multiples of (5)(26) = 130 that are less than 4000. Theanswer is just the integer part of ~ 30.7692. So we grid in 30.

OPTIONAL M ATERIAL CHALLENGE QUESTION

1.Let and be positive integers with j < k . In how many wayscan k be written as a sum of j positive integers?

Solution

Lets begin with some simple examples. Lets try j = 2, k = 3. Then wehave 3 = 1 + 2 = 2 + 1. So there are 2 possibilities.Now lets try j = 3, k = 5. We have5 = 1 + 1 + 3 = 1 + 3 + 1 = 3 + 1 + 1 = 1 + 2 + 2 = 2 + 1 + 2 = 2 + 2 + 1.

Lets think about what we did here. Think of 5 as 1 + 1 + 1 + 1 + 1, andnotice that there are 4 plus signs. We can think of adding two adjacentones as choosing a plus sign. For example if we choose the first plussign we get 2 + 1 + 1 + 1. Its not enough to choose just 1 plus sign. Weneed to choose 2 of them (in this example). If we choose the last twoplus signs we get 1 + 1 + 3. If we choose the first and last plus sign weget 2 + 1 + 2, and so on. In other words we are counting the number ofways to choose 2 of the plus signs from the 4 plus signs. Also, note thatthe number of plus signs is 1 less than 5, and the number we need tochoose is 1 less than 3.

For general j and k with j k, we have k 1 plus signs, and we need tochoose j 1 of them. So the answer is k-1Cj-1 .