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Seven women and nine men are on the faculty in the mathematics department at a school. $\\$ How many ways are there to select a committee of five members of the department if at least one woman and at least one man must be on the committee?

I know the solution and i understood that, we can do it using complement counting answer would be: ${16 \choose 5} - {9\choose 5} - {7 \choose 5}= 4221$

But can anyone tell me, whats wrong with this ? $\color{red} {{9 \choose 1} \times {5 \choose 1} \times {14 \choose 3}}$ ways. First selected exactly 1 man and 1 woman and then rest.

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    Suppose you select A and b initially, and C,d, E later. You could have also selected C,d first and A,b,E later. So there is overcounting. You can't *fractionate* – 2017-01-30
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    Yes, I got the mistake now...thnks :) Even if u have provided this as solution, i could have accepted that :) – 2017-01-30
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    Fine, no sweat ! :) – 2017-01-30

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It counts multiple times the same team.For example it counts as different team the following 2 teams.

1)First pick John , then Mary and then Jane.

2)First pick John, then Jane and then Mary.

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    thnks :) I got it – 2017-01-30