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Math Challenge II-A Combinatorics 10.1 Part b)

 
 
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Re: Math Challenge II-A Combinatorics 10.1 Part b)
LensmireJohn - 2021年10月11日 Monday 11:22
 

For reference, here is the question (2013 AMC 10A Problem 11):

  • 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?

As written, if there are n students, there are $\binom{n}{2}$ ways to pick the welcoming committee and $\binom{n}{3}$ ways to pick the planning committee. This is because members can be on both committees.

If members of the welcoming committee are not allowed to be on the planning committee (and we also assume that the welcoming committee is chosen first), then there are still $\binom{n}{2}$ ways to pick the welcoming committee, but in this case only $\binom{n-2}{3}$ ways to choose the planning committee. (It is n-2 because the two chosen for the welcoming committee are no longer available.

Still assuming there are 10 ways to pick the welcoming committee, $\binom{n}{2} = 10$ so there are $5$ members of the student council. Hence if welcoming committee members cannot be chosen for the planning committee, there would just be $\binom{3}{3} = 1$ way to pick the planning committee.