Question: Okay, this is an example from *Challenge and Trill of Pre-college Mathematics* by Krishnamurthy et al.

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> In how many ways can we form a committee of three from a group of 10 men and 8 women, so that our committee consists of at least one woman?
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I know howit usualy goes: let the answer be *$N$*. The number of committees with no restrictions is $18\choose 3$, and we subtract the number of committees with no women from it, giving
$$N = {{18}\choose{3}} - {{10}\choose{3}} = 816 - 120 = 696$$

However, why is this wrong?

We must choose at least *one* woman, which we can do in $8$ ways. We have to choose two more members, which we can then do in ${17\choose 2}$ ways. Hence
$$N = 8\times {17\choose 2} = 1088 \color{red}{\ne 696} $$

Where's the mistake hiding?

Answer: 
Your first way is absolutely correct, and the easiest way to do it.

Another, more cumbersome, way, is to consider cases: 

1. exactly one woman: ${8 \choose 1}{10 \choose 2} = 8 \times 45 = 360$ ways.
2. exactly two women: ${8 \choose 2}{10 \choose 1} = 28 \times 10 = 280$ ways.
3. exactly three women: ${8 \choose 3} = 56$ ways.

All in all: $360+280+56 = 696$ ways, confirming your own way.

The second way you propose does double counting (which seems logical, as it's more). Namely, whenever there are two or more women in the committtee. If there are, say, exactly two women in the committee, A and B, you can pick A first as one of the 8, and then B as one of the remaining 17. **Or** you can pick B first as one of the eight, and A as one of the 17. So you count that committee twice. For comittees with 3 women, you count them thrice, depending on which one is chosen as one of the 8. If there is one woman only there is no double counting. So you count $360 + 2\times 280 + 3 \times 56 = 1088$..
If you want to use your idea, you'd better separate your discussion into three cases:1) choose 1 woman and two men,2)choose two women and 1 men,3) choose three women

In your answer, you double count some cases, say 1)choose woman A first, then choose woman B, 2)choose woman B and then choose woman A. They are the same combination, but you count them twice