Suppose that a department contains 11 men and 20 women How m

Suppose that a department contains 11 men and 20 women. How many different committees of 6 members are possible if the committee must have strictly more women than men

Solution

possible combinations-        (men,     women)

                                                    (0,6),(1,5),(2,4)

It cannot be 3W and 3 M (this would not be \"strictly\" more women)

Then in each combo, count the ways that each chair can be filled:

In the first case, from 20 women, pick 6
C(20, 6) = 20! / (6! 14!) = 38760

In the second case:
From 20 women pick 5 and from 11 men pick one
C(20,5)*C(11,1) = [20! / (5! 15!) ] * 11=170544

In the third case:
From 20 women pick 4 and from 11 men pick 2
C(20,4)*C(11,2) = [20! / (4! 16!) ] * [11! / (2! 9!) ]=266475

Add all three results.

            38760+170544+266475=475779