6.7 Day 2 Combinations 2011 January 31, 2011
6.7 Day 2
Combinations
Objective: Review permutations and count combinations.
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6.7 Day 2 Combinations 2011 January 31, 2011
Warm­up:
1. In how many ways can you arrange 6 plants on a shelf?
2. In how many ways can 10 dogs line up to be groomed?
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6.7 Day 2 Combinations 2011 January 31, 2011
Review
Permutation: An arrangement of items in a particular order.
P = n r
n!
(n ­ r)!
Example #1: 6P3 METHOD 1
(Formula)
METHOD 2
(Counting Principle)
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6.7 Day 2 Combinations 2011 January 31, 2011
Example #2: Seven yachts enter a race. First, second, and third place trophies will be given to the three fastest yachts. How many arrangements of first, second, and third places are possible with seven yachts?
There are 210 possible arrangements.
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6.7 Day 2 Combinations 2011 January 31, 2011
COMBINATIONS
Combination: Any unordered selection of r objects from a set of n objects.
Cr = n
n!
r!(n ­ r)!
Divide by r! to cancel the repeats.
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6.7 Day 2 Combinations 2011 January 31, 2011
COMBINATIONS
n
Cr = n!
r!(n ­ r)!
Example #3: 7C4 METHOD 1
(Formula)
METHOD 2
(Shortcut)
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6.7 Day 2 Combinations 2011 January 31, 2011
Example #4: Evaluate each expression.
a. 10C5
b. 8C2
c. 25C7
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6.7 Day 2 Combinations 2011 January 31, 2011
Example #5: A reading list for a course in world literature contains 20 books. In how many ways can you choose four books to read?
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6.7 Day 2 Combinations 2011 January 31, 2011
Challenge
Example #6: Ten candidates are running for three seats in the student council. You may vote for as many as three candidates. In how many ways can you vote for three or fewer candidates? 10C3 + 10C2 + 10C1 + 10C0
120 + 45 + 10 + 1 = 176 ways
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6.7 Day 2 Combinations 2011 January 31, 2011
Homework:
page 348 (21 ­ 32, 40, 46 ­ 49, 64)
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