I disagree with both answers.
I'm going to assume that the 3 positions are identical and that it doesn't matter if woman A gets elected to position 1, position 2, or position 3. If I'm wrong with this assumption then there will be a different answer.
Your method allows for repeats. Say the woman are A, B, C, D, and E and the men are V, W, X, Y, and Z. Using your method you first choose one of the five women to get the position reserved for a woman. Let's choose woman B. Now you say choose 2 of the remaining 9 people and put them in the remaining 2 positions. It could be 2 women, 2 men, or 1 of each. So let's choose woman D and man V. So we have B, D, and V elected.
Now try it again. Let's pick woman D to be in the reserved spot. Now we'll choose 2 more people. I pick woman B and man V. So we have D, B, and V selected. But this is the SAME as above.
If you simply select 3 people to be elected out of the 10, you'll get (10 C 3) = 120. So there can't possibly be more than 120 ways to select the 3 people.
The longer way to do this would be to consider each possibility. You can have 3 women and 0 men. You can have 2 women and 1 man. You can have 1 woman and 2 men. That's it. You need at least 1 woman.
For 3 women and 0 men, you have (5 C 3) (5 C 0) = 10.
For 2 women and 1 man, you have (5 C 2) (5 C 1) = 50.
For 1 woman and 2 men, you have (5 C 1) (5 C 2) = 50.
Add them up and you get 110.
The easier way is to realize that the only remaining option is to have 3 men and 0 women. This is just (5 C 3) (5 C 0) = 10. Since we do NOT want this, we take 120 - 10 = 110.
Answer is 110.
May I ask where you are getting your problems from?
