Suppose that a department contains 8 men and 16 women. how many different committees of 6 members are possible if the committee must have strictly more women than men

Begin by writing out all possible ways that women are more than men. 

6 women, 0 men 
5 women, 1 man 
4 women, 2 men <-- 3 men and 3 women would be equal so this is the last one

Now find out the combinations for each (knowing you need to choose 6 people from the group)

(16C6) OR (16C5)(8C1) OR (16C4)(8C2) 
= (16C6) + (16C5)(8C1) + (16C4)(8C2) 
=8008+34944+50960
=93912

Therefore, there should be 93912 ways of arranging this.

Hope I helped :) 

The committees will consist of 5 women and 1 man  OR 4 women and 2 men.

Number of committees of 5 W and 1 M = 16C5 * 5 =  4368
Number with 4 W and 2 M  =  16C4* 5C2 =  1820 * 10 = 18200

Answer is 18200 + 4368 =  22,568