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Archangel Macsika

Suppose that a department contains 8 men and 20 women. How many ways are there to form a committee with 6 members if it must have strictly more women than men?

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Since we have to form a committee of 6 members with strictly more women means that there can be either 4 or 5 or all 6 women in the committee which can be formed in the following number of ways,


8C2*20C4 + 8C1*20C5 + 8C0*20C6


Formula for combination =\frac {n!}{(n-r)!*r!}


Substituting the values of n & r in the above formula we get,


8C2*20C4 + 8C1*20C5 + 8C0*20C=298452


Hence we can form a committee of 6 members in 298452 ways if it must have strictly more women than men


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Question ID: mtid-5-stid-8-sqid-3624-qpid-2323