Difference between revisions of "2013 AMC 10A Problems/Problem 11"
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If there are <math>5</math> people on the welcoming committee, then there are <math>\dbinom{5}{3} = 10</math> ways to choose a three-person committee, <math>\textbf{(A)}</math>. | If there are <math>5</math> people on the welcoming committee, then there are <math>\dbinom{5}{3} = 10</math> ways to choose a three-person committee, <math>\textbf{(A)}</math>. | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2013|ab=A|num-b=10|num-a=12}} | ||
Revision as of 21:08, 7 February 2013
Problem
A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?
Solution
Let the number of students on the council be 
. We know that there are 
ways to choose a two person team. This gives that 
, which has a positive integer solution of 
.
If there are people on the welcoming committee, then there are 
ways to choose a three-person committee, 
.
See Also
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
