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I encountered a math question yesterday. It read:

There are 10 men and 6 women in a group. Out of that group, a committee of five should be chosen. The committee must contain at least 3 men. Find all the possible combinations for this committee.

I know the answer to this is supposed to be:

$$ \begin{pmatrix} 10\\ 3\\ \end{pmatrix} * \begin{pmatrix} 6\\ 2\\ \end{pmatrix} + \begin{pmatrix} 10\\ 4\\ \end{pmatrix} * \begin{pmatrix} 6\\ 1\\ \end{pmatrix} + \begin{pmatrix} 10\\ 5\\ \end{pmatrix} * \begin{pmatrix} 6\\ 0\\ \end{pmatrix} $$

However, I was wondering why the following logic doesn't work: $$ \begin{pmatrix} 10\\ 3\\ \end{pmatrix} * \begin{pmatrix} 13\\ 2\\ \end{pmatrix} $$ This is what I thought the solution would be. I would first get all the combinations possible of the first initial 3 men chosen, then I get all the combinations of the possible last two members of the committee out of the 13 people left in the group. It seems like sound logic to me, so I'm not sure why it doesn't give the correct answer. My friend told me it's because I'm including the men twice in my combination calculations, which causes excessive combinations to become part of my answer.

Could anyone please clear this up for me?

Andy
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1 Answers1

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Your answer somehow treats the three men picked in the first step to be "special" compared to any additional men chosen in the second step. Letting capital letters represent men and lowercase women, the result of (ABC)(Da) is the "same" result as (ABD)(Ca) since in both results you have the men ABCD and the woman a, yet your count treated these cases as being different and counted each individually.

You should avoid errors caused by putting unnecessary "importance" on people based on at what point in the process they are picked.

JMoravitz
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