Let's call the women A, B, C, D, E, F, G, H, I and the men S, T, U, V, W, X, Y, Z. If you are going to choose $5$ of the women and then $7$ of the rest, here is one way to do it:
(1) choose the women A, B, C, D, E and then choose Z, Y, X, W, V, U, I.
Here is another way:
(2) choose the women A, B, C, D, I and then choose Z, Y, X, W, V, U, E.
But actually the results of these two choices are the same. So you have counted this possibility twice - in fact, if you think about it, more than twice. If you do all the calculations you should find that your second answer is greater than the correct answer.
A) 3 of 10 is ${10 \choose 3}=\frac{10.9.8}{3.2.1}=10.3.4=120$.
B) 3 of 6 is ${6 \choose 3}=\frac{6.5.4}{3.2.1}=5.4=20$.
C) The probability that a committee is of 3 men is 20/120=1/6.
D) Two men and one woman can be chosen in ${6 \choose 2}{4 \choose 1}=\frac{6.5}
{2.1}\frac{4}{1}=15.4=60$.
E) The probability is 60/120=1/2.
Best Answer
From all 13C6 solutions, you must subtract the solutions where both A and B are chosen.
If you choose both A and B, there are 11C4 possibilities for the other 4 in the committee.
So 13C6 - 11C4 = 1386.
You calculated the committees where it was not allowed to have exactly one of A and B.