Difference between revisions of "2013 AMC 10A Problems/Problem 11"
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Let the number of students on the council be <math>x</math>. We know that there are <math>\dbinom{x}{2}</math> ways to choose a two person team. This gives that <math>x(x-1) = 20</math>, which has a positive integer solution of <math>5</math>. | Let the number of students on the council be <math>x</math>. We know that there are <math>\dbinom{x}{2}</math> ways to choose a two person team. This gives that <math>x(x-1) = 20</math>, which has a positive integer solution of <math>5</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> | + | 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>. |
Revision as of 18:58, 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, 
.
