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Question: There are $10$ men and $12$ women. How many ways are there to choose a group of $5$ people where at least $2$ are women?

The correct answer is $${12 \choose 2} * {10 \choose 3} + {12 \choose 3} * {10 \choose 2} + {12 \choose 4} * {10 \choose 1} + {12 \choose 5} * {10 \choose 0}$$ which is pretty straight-forward and correct.

But say we chose $2$ women first, and chose the rest $3$ from the remaining people. So the combination becomes ${12 \choose 2} * {20 \choose 3}$. But it doesn't match with the first answer. What am I doing wrong in the second process?

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    $\begingroup$ Side note: it's easier to compute the number of combinations with no women or exactly one woman. then subtract from the total. $\endgroup$ – lulu Apr 22 '17 at 12:04
  • $\begingroup$ @michaelhowes It's clear now. Thanks! $\endgroup$ – J. R. Apr 22 '17 at 12:07
  • $\begingroup$ @michaelhowes I suggest you transform your comment into an answer. $\endgroup$ – N. F. Taussig Apr 23 '17 at 0:46
  • $\begingroup$ @N.F.Taussig I have followed your suggestion. $\endgroup$ – michaelhowes Apr 23 '17 at 4:43
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The problem with your second approach is that you are over counting.

For instance, if the group you pick is one with three women, you would count the same group 3 times.

To see this let the people you pick be $A,B,C,D,E$ where $A,B,C$ are the three women. You could have $A,B$ be the women you first choose and then have $C,D,E$ by the people you choose from the rest. Or A,C could be the first women and $B,D,E$ be the rest. Or $B,C$ be the first women and $A,D,E$ be the rest.

In the first answer this group is correctly only counted once.

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You are over counting. The thing is every time you choose two women, and then 3 other members. You are also counting the possibility of having say 4 women, 2 of them, counted in the first term and 2 counted in second. You will have counted this case again when vice versa is true, the latter 2 women counted in the first term and first 2 women counted in the second term (and keeping the man same). So you have counted this case twice, there are total of 6 ways to count, as pointed in the comments, and so there will be several other cases counted more than once.

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    $\begingroup$ The calculation $\binom{12}{2}\binom{20}{3}$ actually counts the case in which four women are selected $\binom{4}{2} = 6$ times, once for each way of choosing two of those four women first. $\endgroup$ – N. F. Taussig Apr 22 '17 at 19:25
  • $\begingroup$ Oh alright, that sure makes sense :D $\endgroup$ – Rakesh Arya Apr 22 '17 at 20:03
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You can think of ${12\choose2}{20\choose3}$ as counting the number of ways to choose a $5$-member committee with two co-leaders, both of whom must be women. The fact that you are distinguishing two members of each committee means that your count will be larger than simply counting (leaderless) committees, because committees with more than two women have more than one way to have co-leaders.

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