Chapter 10
Lesson
10-1
Combinations
Vocabulary
combination
number of combinations of n
things taken r at a time,
n
nC r, ( 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.
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?
Activity
Sierra is making trail mix for a hike with friends. She has five available
ingredients: granola, peanuts, raisins, sunflower 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 five 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 first 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 find?
On the trail Trail mix, a
nutritious, high energy, and
lightweight snack eaten while
hiking
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.
618
Binomial Distributions
SMP_SEFST_C10_L01_618_623_FINAL.618 618
6/9/09 2:49:30 PM
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
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
5!
choosing three ingredients in order is 5P 3 = _
or 5 · 4 · 3. In Step 3,
2!
the number of ways of reordering any set of three ingredients is
3P 3 = 3!. The number 3! is the number of times each trail mix was
counted in Step 1. Thus, to find the number of different unordered
mixes, you can divide the number of ordered mixes by the number of
times each unordered mix appears.
( 2! ) _
5·4
5P 3
_
=_
= 5! = _
= 10
5!
_
3P 3
3!
3!2!
2
The number of combinations of 5 ingredients taken 3 at a time is
written 5C 3. In general, the number of combinations of n things taken r
at a time is written nCr or ( nr ). 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.
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.
Theorem (Formula for nCr)
nP r
n!
For all integers n and r, with 0 ≤ r ≤ n, nCr = _
=_
.
r!
(n - r)!r!
Proof Any combination of r objects can be arranged in r! ways. So r! · nCr = nPr .
1
_
Solve for nCr .
nCr = nPr · r!
Substitute from the Formula for nPr Theorem.
Thus,
QY2
n!
1
n!
_
_
·_
r! = (n - r)!r! .
nCr = (n - r)
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.
GUIDED
Example 1
A 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?
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.
(continued on the next page)
Combinations
SMP_SEFST_C10_L01_618_623_FINAL.619 619
619
6/9/09 2:50:50 PM
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.
10!
10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1
C
= _ = ___
10 6
4!6!
4 · 3 · 2 · 1 ·6 · 5 · 4 · 3 · 2 · 1
10 · 9 · 8 · 7
= __ = 10 · 3 · 7 = 210
4·3·2·1
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
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 five 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 five
tosses. To answer this question, think of the pennies as occupying five
positions along a line, numbered 1 through 5.
1
2
3
4
How many different ways
are there of ordering
7 main courses from
a list of 10?
5
Each way of getting three tails is like picking three of the five 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 five tosses is the number of combinations of five things taken
three at a time, 5C 3.
QY4
Each time you get exactly three tails in five tosses, the other two coins
are heads. So the number of ways of getting three tails in five tosses is
the same as the number of ways of getting two heads in five tosses. That
is the idea behind the two solutions to Example 2.
620
QY4
How many ways are there
to get 3 heads on a toss
of 7 coins?
Binomial Distributions
SMP_SEFST_C10_L01_618_623_FINAL.620 620
6/9/09 2:51:20 PM
Lesson 10-1
Example 2
A 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?
30!
C = _ = 142,506
30 25
5!25!
Solution 2 Think: How many ways are there to get exactly 5 out of
30 questions wrong?
30!
C = _ = 142,506
30 5
25!5!
“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 3
A 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
C ·
C
3 3 247 22
__
250C25
247!
1· _
(
)
225!22!
247! · 25 · 24 · 23 · 22!
247!25!
=_
= _ = ___
250!
(_
225!25! )
250!22!
1
250 · 249 · 248 · 247! · 22!
3
25 · 24 · 23
= __
≈ 0.000894.
250 · 249 · 248
10
31
Combinations
SMP_SEFST_C10_L01_618_623_FINAL.621 621
621
6/16/09 10:21:36 AM
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 five girls
out of seven children
5. Match each item on the left with one on the right.
a. combination
b. permutation
i. an arrangement of objects in which
different orderings are distinguishable
ii. an arrangement of objects in which
different orderings are equivalent
6. a. How many combinations of the letters UNESCO,
taken five 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.
( 284 )
8.
34C 1
The United Nations Educational,
Scientific and Cultural Organization
(UNESCO) Secretariat Building in
Paris, France
9. a. How many ways can you flip 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 flips?
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.
622
( n1 )
12.
nCn-1
13.
nC n
Binomial Distributions
SMP_SEFST_C10_L01_618_623_FINAL.622 622
6/9/09 2:52:53 PM
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.
13P5
16. C = _
x
5
5!
17. 7! · xC 7 = 31P 7
18.
( x -x 2 ) = 325
19. A lottery picks five 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 first 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 five 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 first-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
XY YXXX. (Hint: Determine the number of 4-game series, 5game 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.
QY ANSWERS
1. a. permutations
b. combinations
9!
9P4
_
2. 9C4 = _
4! = 5!4!
3. 120
4. 7C3 = 35
Combinations
SMP_SEFST_C10_L01_618_623_FINAL.623 623
623
6/15/09 10:39:31 AM