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 the same number of men and women?
If it must be the same number of men and women, it should be 3 men and 3 women in the department.
So we can choose 3 men out of 10 in "C^3_{10}=\\frac{10!}{3!7!}=120" ways.
We can choose 3 women out of 15 women in "C^3_{15}=\\frac{15!}{12!3!}=455" ways.
So we can form the department in 120x455=54600 ways.
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