There was a problem discussed in an another post by

12C3*9C2 + 12C2*9C3 + 12C1*9C4 + 12C0*9C5 = 7920+5544+1512+126 = 15102 possible ways
Possible ways of forming the 5 person committee given that there must be 3 women or less:
I) 3W+2M
II) 2W+3M
III) 1W+4M
IV) 0W+5M
The sum of these possibilities will give you the total possible favorable outcomes. Since there are 12 women from which to select the women in the committee and 9 men from which to select the men in the committee, we have the following:
12C3 ways of selecting 3 women * 9C2 ways of selecting other 2 men
12C2 ways of selecting 2 women * 9C3 ways of selecting other 3 men
12C1 ways of selecting 1 woman * 9C4 ways of selecting other 4 men
12C0 ways of selecting 0 woman * 9C5 ways of selecting other 5 men
The answer is the sum of those combination outcomes and as shown in first line for question 3.

21C5 - 9C4*12C1 - 9C5 = 20349 - 1512 -126 = 18711.