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?
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Since there must be more women, there are 3 cases to consider. 6 women, 0 men 5 women, 1 men 4 women, 2 men $\displaystyle \sum_{k=0}^{2}C(15,6-k)C(10,k)$ C(15,6)C(10,0)+C(15,5)C(10,1)+C(15,4)C(10,2)
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