Page 820 - Algebra 1

In Example 1, there are 6 times as many permutations as combinations. For each set of 3 letters, there are 3 · 2 · 1 = 6 different ways to order the letters. To find the number of combinations, nCr, when selecting r out of n items, divide the number of permutations, nPr, by the number of way_s to order r items, r!.
n! _____
That is, nCr = nPr = (n - r)! = n! . number of ways to order r items r! r!(n - r)!
Math Language
nCr is read “n choose r.” So, 4C3 is read, “Four choose three.”
Combination Formula
The number of combinations of n_items taken r at a time is
nCr = n! . r!(n - r)!
Example
2
Finding the Number of Combinations
a. At a restaurant 2 side dishes may be chosen. There are a total of 6 side dish choices. How many combinations are there?
SOLUT_ION
n! nCr = _
Use the combination formula. Substituten=6andr=2. Simplify inside parentheses. Simplify.
r!(n - r)! 6C2 = 6!
2!(6 - 2)! = 6!
_ _2!4!
= 720 = 15 2 · 24
There are 15 ways to choose 2 side dishes.
b. A company delivers fruit to its customers every month. There are
16 different types of fruit. Each customer can choose 12 types of fruit each year. How many combinations can each customer make?
Use the combination formula. Substitute 16 for n and 12 for r. Simplify inside parentheses.
Rewrite 16! as 16 · 15 · 14 · 13 · 12!. Cancel 12!.
SOLUTI_ON nCr = n!
r!(n - r)! __
16C12 = 16! _12!(16 - 12)!
= 16! 12!4!
= 16 · 15 · 14 · 13 · 12! __
= 16 · 15 · 14 · 13 __
12!4!
4!
There are 1820 ways to choose 12 fruits.
Online Connection www.SaxonMathResources.com
= 1820
Simplify.
Lesson 118 805