1. ## combinational probability

suppose that a department contains 10 men and 15 women. How many ways are there to form a committee with six members if it must have more women than men?

2. Since there must be more women, there are 3 cases to consider.

6 women, 0 men
5 women, 1 men
4 women, 2 men

$\sum_{k=0}^{2}C(15,6-k)C(10,k)$

C(15,6)C(10,0)+C(15,5)C(10,1)+C(15,4)C(10,2)