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Consider the number of ways to choose a committee of $3$ members from $5$ men and $2$ women where there must be at least $1$ woman.

First, we can choose a woman for the committee which can be done in $2 \choose 1$ ways and then we can choose the rest from the 6 people left in $6 \choose 2$ ways which gives us $2 \choose 1$$6 \choose 2$ $= 30$ ways.

But if we divide it into cases: first if we consider the case where there is only $1$ woman and $2$ men in the committee, we get $2 \choose 1$$5 \choose 2$ $= 20$ ways. Then we consider the case where there are $2$ women and $1$ man in the committee. Here we get $2 \choose 2$$5 \choose 1$ $= 5$ ways. Summing up we get a total of $25$ ways.

But why are the processes giving us different answers? What am I missing?

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    $\begingroup$ Your first method counts those committees with two women twice. $\endgroup$ – lulu Feb 12 at 11:48
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Your first method counts those committees with two women twice.

More specifically: choosing $W_1$ as the first woman and then $M_1,W_2$ as the other two results in the same committee as choosing $W_2$ as the first woman and then $M_1,W_1$ as the other two, but your first method counts them as two different solutions.

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