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? Solution The committee may contain (1) 6 women, or (2) 5 women and 1 man, or (3) 4 women and 2 men. Now (1) can be done in C(15,6) ways, (2) can be done in C(15,5)*C(10,1) ways, and (3) can be done in C(15,4)*C(10,2) ways. Add up the number of ways to do (1) + (2) + (3). Note: C(a,b) = a!/(b!*(a-b)!)..