target780 wrote:

What is wrong with this approach:

first select 1 man. it can be done in 5C1 = 5 ways.

now there are 7 people left (4 M and 3 W) .. select any 2 out of the 7, it can be done is 7C2 = 21 ways

so the total number is 21 * 5 = 105.

This is a conundrum. I've been trying to work out why your way doesn't work, and I know why, but I don't know if I can explain it well.

5x7x6/2 assumes that the first space is only one guy, and the other spaces can be anything, thereby encompassing all other sorts of combos, with women only, some men and some women, and no women, just men. But what it also does is it allows doubling of some combinations. For example, imagine that we have our five men a, b, c, d, and e, and our 3 women f, g, and h.

Now here is one combination of 2 men and 1 woman:

b, d, g

Another combination would be

d, b, g

But those are the same thing, and should only be counted once. Using your method, they are counted twice.

In the correct method, we would figure out how many ways 2 men and 1 woman can come together, and we would automatically elminate the second possibility.

That's not the most complete answer, but I hope it sheds some light on things. I actually drew a very involved diagram to understand it all - the tree diagram. If you do the same, it should make sense.