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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>.
  
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If there are <math>5</math> people on the welcoming committee, then there are <math>\dbinom{5}{3} = \boxed{\textbf{(A) }10}</math> ways to choose a three-person committee.
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If there are <math>5</math> people on the student council, then there are <math>\dbinom{5}{3} = \boxed{\textbf{(A) }10}</math> ways to choose a three-person committee.
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==See Also==   
 
==See Also==   
 
{{AMC10 box|year=2013|ab=A|num-b=10|num-a=12}}
 
{{AMC10 box|year=2013|ab=A|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:18, 24 December 2017

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?


$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25$

Solution

Let the number of students on the council be $x$. We know that there are $\dbinom{x}{2}$ ways to choose a two person team. This gives that $x(x-1) = 20$, which has a positive integer solution of $5$.

If there are $5$ people on the student council, then there are $\dbinom{5}{3} = \boxed{\textbf{(A) }10}$ ways to choose a three-person committee.

See Also

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AMC 10 Problems and Solutions

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