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2013 AMC 10A Problem 11

Problem 11 of 25IntermediateAlgebraCounting & Probability

A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 1010 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?

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Solution

Let xx be the number of students. Then the number of ways to pick a two-person committee is (x2)=x(x−1)2. \binom{x}{2} = \dfrac{x(x - 1)}{2}. We know that this equals 10,10, so x2−x=20 x^2 - x = 20 x2−x−20=0. x^2 - x - 20 = 0. Factoring yields (x−5)(x+4)=0 (x - 5)(x + 4) = 0 x=5, x = 5, since there cannot be a negative number of students. Then, the number of ways to pick a 33-person committee is (53)=(52)=10. \binom{5}{3} = \binom{5}{2} = 10. Thus, A is the correct answer.

More practice

Concepts: combinations · quadratic

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