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Ok, So today in my math class we were going over permutations and combinations. The problem was a fairly normal one of if you have seven women and nine men, how many ways can you do a group of five that contains at least one woman.

So I said, ok, so our first place has to be a woman, and there are seven of them then for the remaining 4 it does not matter who we pick. So I wound up with $$ \frac{7*15*14*13*12}{5!} = 1911 $$ as the answer. The teacher did the normal method of $$ \binom{16}{5} - \binom{9}{5} = 4242 $$ I understand why the teacher is right, but I don't understand why I'm wrong. Thank you.

Henrik supports the community
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Alonzo Muncy
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    Your method would appear to yield $7\times \binom {15}4= 9555$, no? Regardless, your method overcounts. Take a group of five women...your method counts that one five times. – lulu Jul 06 '17 at 15:16
  • I divided by 5! not 4!. – Alonzo Muncy Jul 06 '17 at 15:19
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    I see that, but I don't understand why. I thought your method was "select a woman, $7$ choices there. Then select $4$ more from the remaining $15$ people. $\binom {15}4$ choices." I don't understand where the extra $5$ in the denominator comes from. – lulu Jul 06 '17 at 15:21
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    One good way to analyze a counting method is to do it with smaller numbers. Say I have two women $W_1,W_2$ and two men $M_1,M_2$ and that I want to form a two person group with at least one woman. What does your method yield? – lulu Jul 06 '17 at 15:22
  • https://math.stackexchange.com/questions/2340044/in-combinatorics-how-can-one-verify-that-one-has-counted-correctly –  Jul 06 '17 at 15:24

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Your problem occurs at the point when you divide out by $5!$. The usual justification for a step like this, which I presume you've seen in other problems, goes something like: "I've counted different permutations like $ABCDE$, $DECAB$ (etc), separately in the first step, but they give the same group. Each group has $5!$ permutations, so I've counted each group $5!$ times, so dividing that number by $5!$ gives the right number of groups."

Let's see how this fails in your method. Consider a selection like $W_1M_1M_2M_3M_4$: when you divide out by $5!$ you assume that you've (over-)counted other permutations of it like $M_3M_4W_1M_1M_2$... but you wouldn't have, because your counting method always required that the first chosen person be a woman. Most of the permutations would not have been counted by your method.

In theory, you could try to repair your method by checking the number of permutations that leaves a woman in first place, but that would depend on the number of women in the group and probably turns into a really complicated sum.

Chessanator
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