Difference between revisions of "2007 AMC 10A Problems/Problem 12"
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== Solution == | == Solution == | ||
− | Each tourist has to pick in between the <math>2</math> guides, so for <math>6</math> tourists there are <math>2^6</math> possible groupings. However, since each guide must take at least one tourist, we subtract the <math>2</math> cases where a guide has no tourist. Thus the answer is <math>2^6 - 2 = 62\ \mathrm{(D)}</math>. | + | Each tourist has to pick in between the <math>2</math> guides, so for <math>6</math> tourists there are <math>2^6</math> possible groupings. However, since each guide must take at least one tourist, we subtract the <math>2</math> cases where a guide has no tourist. Thus the answer is <math>2^6 - 2 = \boxed{62}\ \mathrm{(D)}</math>. |
== See also == | == See also == |
Revision as of 19:16, 31 October 2020
Problem
Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible?
Solution
Each tourist has to pick in between the guides, so for tourists there are possible groupings. However, since each guide must take at least one tourist, we subtract the cases where a guide has no tourist. Thus the answer is .
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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