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618 Binomial Distributions

Lesson

Combinations

Chapter 10

Vocabulary

combination

number of combinations of n

things taken r at a time,

nC

r, ( n r )

BIG IDEA With a set of n elements, it is often useful to be able to compute the number of subsets of size r.

In earlier lessons, you found answers to counting problems involving permutations. Recall that a permutation of r of n items is an ordering of r objects taken from the n objects. Numbers of permutations can be found using the Multiplication Counting Principle from Lesson 6-2.

In contrast, a combination is a collection of objects in which order does not matter. For instance, in making trail mix, if ingredients are added in different orders, the end result is the same. The Activity exhibits a process that can be used to determine numbers of combinations.

Sierra is making trail mix for a hike with friends. She has fi ve available

ingredients: granola, peanuts, raisins, sunfl ower seeds, and dried blueberries.

Knowing that her friends have different food preferences, she decides to make

small bags of mix, each with a different selection of three ingredients, rather

than a big bag with all fi ve ingredients. How many bags can she make so that

each bag has a different mix?

Step 1 Using the Multiplication Counting Principle, Sierra reasons that there are 5 choices for the fi rst ingredient in the mix, 4 choices for the second ingredient, and 3 choices for the third. How many mixes does she think she can make?

Step 2 Worried that she is going to lose track of what was put in each bag, Sierra makes labels. Her label GPB means granola-peanuts-blueberries. Make a list of the possibilities. Make sure that you avoid duplicate mixes. BGP is the same mix as GPB. Check your list with another student’s list to make sure that you have listed all the possibilities. How many did you fi nd?

Step 3 How should Sierra adjust her answer from Step 1 to produce the number of mixes you listed in Step 2? (Hint: If you have chosen three ingredients, say granola-peanuts-blueberries, how many different ways could you write the three-letter label?)

Step 4 Repeat Steps 1–3 as if Sierra had six ingredients instead of 5. Check your answer by listing all the possibilities.

ActivityActivity

On the trail Trail mix, a

nutritious, high energy, and

lightweight snack eaten while

hiking

On the trail Trail mix, a

nutritious, high energy, and

lightweight snack eaten while

hiking

10-1

Mental Math

In the 26-letter English

alphabet, how many

different 2-letter initials

are possible

a. if the same letter can be repeated?

b. if the same letter cannot be repeated?

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Combinations 619

Lesson 10-1

A Formula for Counting Combinations

Step 1 of the Activity asks for the number of permutations of 3 of 5 things. Step 2, by contrast, asks for the number of combinations of 5 things taken 3 at a time. Notice the difference between permutations and combinations. In permutations, order matters and so different orders are counted as unique. In combinations, BGP and GPB are considered the same. They are different permutations of 3 letters but they are the same combination of 3 letters.

QY1

The Activity shows how to compute a number of combinations by dividing two numbers of permutations. In Step 1, the number of ways of choosing three ingredients in order is 5P3 = 5! _ 2! or 5 · 4 · 3. In Step 3, the number of ways of reordering any set of three ingredients is

3P3 = 3!. The number 3! is the number of times each trail mix was counted in Step 1. Thus, to fi nd the number of different unordered mixes, you can divide the number of ordered mixes by the number of times each unordered mix appears.

5P3 _

3P3 =

( 5! _ 2!

) _

3! = 5! _ 3!2! = 5 · 4 _ 2 = 10

The number of combinations of 5 ingredients taken 3 at a time is written 5C3. In general, the number of combinations of n things taken r

at a time is written nC

r or ( n r ) . Some people read this as “n choose r.” A

formula for nCr can be derived from the formula for nPr by generalizing the argument above.

Theorem (Formula for nC

r)

For all integers n and r, with 0 ≤ r ≤ n, nC

r = n

Pr _ r! = n! _ (n - r)!r! .

Proof Any combination of r objects can be arranged in r! ways. So r! · nC

r =

nP

r .

Solve for nC

r.

nC

r =

nP

r · 1 _ r!

Substitute from the Formula for nP

r Theorem.

Thus, n

Cr = n!

_ (n - r) · 1 _

r! = n! _ (n - r)!r! .

QY2

Because order does not matter in counting combinations, you can consider nCr as the number of subsets of size r of a set of n objects.

Example 1A restaurant menu has ten appetizers and twelve main courses. A group of

people decides they want to order six appetizers and four main courses. In

how many ways can they do this?

(continued on the next page)

QY1

Which situation involves combinations? Which situation involves permutations?

a. Three scholarships worth $1500, $1000, and $500 are distributed to students from a class of 800.

b. Three $1000 scholarships are distributed to the students of a class of 800.

QY1

Which situation involves combinations? Which situation involves permutations?

a. Three scholarships worth $1500, $1000, and $500 are distributed to students from a class of 800.

b. Three $1000 scholarships are distributed to the students of a class of 800.

QY2

Write the formula for the number of ways of making bags of 4-ingredient trail mix if 9 ingredients are available. Do not compute an answer.

QY2

Write the formula for the number of ways of making bags of 4-ingredient trail mix if 9 ingredients are available. Do not compute an answer.

GUIDEDGUIDED



Chapter 10

Solution First consider the appetizers. There are 10 appetizers, from which 6 are to be chosen. This is like choosing a subset of 6 objects from a set of 10 objects, so use the formula for 10C6.

10C6 = 10! _ 4!6! = 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 ___ 4 · 3 · 2 · 1 · 6 · 5 · 4 · 3 · 2 · 1

= 10 · 9 · 8 · 7 __ 4 · 3 · 2 · 1 = 10 · 3 · 7 = 210

For the main courses, there are 12 dishes, from which ? are to be chosen. Since all the courses come out at once, use the formula for ? C ? to get

12! _ ? ! ? !

= ? total main course combinations.

To compute the total number of meals, use the Multiplication Counting Principle: 210 appetizer combinations · ? main course combinations =

? different meals.

Check Use a calculator.

QY3

Using Combinations to Count Arrangements

As you will see in this chapter, combinations are very important in the calculation of certain probabilities. For instance, suppose fi ve pennies are tossed. If you want to know the probability of getting three tails, then you need to know how many ways three tails can occur in fi ve tosses. To answer this question, think of the pennies as occupying fi ve positions along a line, numbered 1 through 5.

Each way of getting three tails is like picking three of the fi ve numbers. For instance, if coins 3, 4, and 5 are tails, as pictured above, then you have picked the numbers 3, 4, and 5. Do you see that this is just like Sierra’s problem from the Activity? The number of ways of getting three tails in fi ve tosses is the number of combinations of fi ve things taken three at a time, 5C3.

QY4

Each time you get exactly three tails in fi ve tosses, the other two coins are heads. So the number of ways of getting three tails in fi ve tosses is the same as the number of ways of getting two heads in fi ve tosses. That is the idea behind the two solutions to Example 2.

QY3

How many different ways are there of ordering 7 main courses from a list of 10?

QY3

How many different ways are there of ordering 7 main courses from a list of 10?

1 2 3 4 5 1 2 3 4 5

QY4

How many ways are there to get 3 heads on a toss of 7 coins?

QY4

How many ways are there to get 3 heads on a toss of 7 coins?


Combinations 621

Lesson 10-1

Example 2A vocabulary test has 30 items. Each correct answer is worth 1 point and

each wrong answer is worth 0 points. How many ways are there of scoring 25

points?

Solution 1 Think: How many ways are there of selecting 25 questions out of 30 to be correct?

30C25 = 30! _ 5!25! = 142,506

Solution 2 Think: How many ways are there to get exactly 5 out of 30 questions wrong?

30C5 = 30! _ 25!5! = 142,506

“Thinking Factorially”:Computing Combinations by Hand

Sometimes numbers resulting from computing a combination are too large for a calculator’s memory or display. Even when a calculator can compute and display the exact number of combinations, it can be easier and more illuminating to verify the calculator’s results by doing some computations by hand. When working with quotients of factorials, common pairs of factors appear and cancel each other. The result is a more manageable fraction. We describe this method of simplifying calculations as “thinking factorially.”

Example 3A school requires that all participants in sports submit to random drug testing.

During one round of tests, 25 students are randomly selected from among

the 250 students participating in sports. What is the probability that the

quarterback of the football team and the goalies of both the boys’ and girls’

soccer teams are three of the students selected?

Solution There are 250C25 combinations of 25 players that could be picked. This number is huge, about 1.65 · 1034. This number is the denominator of the probability to be calculated. The numerator is the number of these 1.65 · 1034 combinations of 25 players that include the three given players. Notice that each such combination includes 22 players from the remaining 247 athletes. There is only one way to select the three given players (3C3 = 1) and there are 247C22 ways to select the other 22 players. Thus, the probability that these three players are selected, assuming randomness, is

3C3 · 247C22 __ 250C25

= 1 · ( 247! _ 225!22! ) _

( 250! _ 225!25! ) = 247!25! _ 250!22! = 247! · 25 · 24 · 23 · 22! ___ 250 · 249 · 248 · 247! · 22!

= 25 · 24 · 23 __

250 · 249 · 248 ≈ 0.000894.1 3

10 31

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Chapter 10

Questions

COVERING THE IDEAS

In 1–4, tell whether the situation uses permutations or combinations.

You do not have to compute an answer.

1. number of ways to assign 25 students to 30 desks 2. number of committees of 5 students out of a class of 30 students 3. number of ways of picking 3 books out of a reading list of 10 books 4. number of different orders in which a family could have fi ve girls

out of seven children

5. Match each item on the left with one on the right. a. combination i. an arrangement of objects in which

different orderings are distinguishable b. permutation ii. an arrangement of objects in which

different orderings are equivalent

6. a. How many combinations of the letters UNESCO, taken fi ve at a time, are possible?

b. Explain why the result of Part a is the same as the number of combinations taken one at a time.

c. Which would be easier to list, the combinations in Part a or those in Part b?

d. Explain what would be different if you were asked to count all 5-letter strings from UNESCO.

In 7 and 8, evaluate.

7. ( 28 4 ) 8. 34C1

9. a. How many ways can you fl ip 10 coins and get exactly 3 heads? b. Using the Multiplication Counting Principle, how many total

outcomes are possible? c. Using your answers from Parts a and b, what is the probability

of getting three heads in ten fl ips?

10. Suppose all 4 members of the school golf team are among the 30 athletes chosen for drug testing at a school with 150 athletes.

a. Write an expression for the number of combinations of players that could be chosen.

b. Write an expression for the number of combinations that include the four golf team members.

c. What is the probability that all 4 golf team members are chosen?

APPLYING THE MATHEMATICS

In 11–13, evaluate the expression. Then explain your answer in terms

of choosing objects from a collection.

11. ( n 1 ) 12. nCn-1 13. nCn

The United Nations Educational,

Scientifi c and Cultural Organization

(UNESCO) Secretariat Building in

Paris, France

The United Nations Educational,

Scientifi c and Cultural Organization

(UNESCO) Secretariat Building in

Paris, France


Combinations 623

Lesson 10-1

14. Pete’s A Pizza restaurant menu lists fi fteen different toppings. a. How many different pizzas could be ordered with six toppings? b. How many different pizzas could be ordered with nine toppings?

15. Give an example of a situation that would lead to the number 100C10.

In 16–18, solve for x.

16. xC5 = 13P5 _ 5! 17. 7! · xC7 = 31P7 18. ( x x - 2 ) = 325

19. A lottery picks fi ve balls from a bin of 55 white balls labeled 1 to 55 and one ball from a bin of 42 green balls labeled 101 to 142. To win the jackpot, a participant must correctly guess all six numbers, although the order of the white numbers is irrelevant. Tickets cost $1. If someone were to buy all possible tickets, how much would it cost?

20. Six points are in a plane with no three of them collinear, as shown. a. How many triangles can be formed having these points

as vertices? b. Generalize your result in Part a to the case of n points.

REVIEW

21. A theater has 17 seats in the fi rst row and there are eight more seats in each subsequent row. If the last row has 217 seats, how many rows are there in the theater? (Lesson 8-1)

22. Solve for n: 23! = n · 21! (Lesson 6-4)

23. When one coin is tossed there are two possible outcomes—heads or tails. How many outcomes are possible when the following numbers of coins are tossed? (Lesson 6-3)

a. 2 b. 5 c. 8 d. n

In 24 and 25, two bags each contain fi ve slips of paper. An angle is

drawn on each slip. The measures of the angles in each bag are 10°,

30°, 45°, 60°, and 90°. Let a and b be the measures of the angles

pulled from bags 1 and 2, respectively. (Lessons 6-1, 4-2)

24. Find P(a + b ≥ 90°). 25. Find P(sin(a + b) ≥ sin 90°).

EXPLORATION

26. In professional baseball, basketball, and ice hockey, the champion is determined by two teams playing a fi rst-to-four wins out of seven games series. Call the two teams X and Y.

a. In how many different ways can the series occur if team X wins? For instance, two different 6-game series are XXYXYX and XYYXXX. (Hint: Determine the number of 4-game series, 5-game series, 6-game series, and 7-game series.)

b. There is an nCr with n and r both less than 10 which equals the answer for Part a. Find n and r. c. Explain why the combination of Part b answers Part a.

The Stanley Cup, pictured

here, is given to the champion

of the National Hockey

League after a series like that

in Question 26.

The Stanley Cup, pictured

here, is given to the champion

of the National Hockey

League after a series like that

in Question 26.

QY ANSWERS

1. a. permutations

b. combinations

2. 9C4 = 9P4 _ 4! = 9!

_ 5!4!

3. 120

4. 7C3 = 35

QY ANSWERS

1. a. permutations

b. combinations

2. 9C4 = 9P4 _ 4! = 9!

_ 5!4!

3. 120

4. 7C3 = 35

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