Suppose that a department contains 8 men and 20 women. How many ways are there to form a committee with 6 members if it must have strictly more women than men?

Answer 1

If more women than men, possibilities are
6W, 0M
5W, 1M
4W, 2M
3W, 3M --- no longer possible

6W,0M ---> C(20,6) x C(8,0) = 38760
5W, 1M ---> C(20,5) x (8,1) = 15504(8) = 124032
4W, 2M ---> C(20,4) x C(8,2) = 4845(28) = 135660

add them up