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15 January, 17:21

How many possible committees of 4 people can be chosen from 15 men and 20 women so that at least two women should be on each committee?

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  1. 15 January, 17:24
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    If there has to be 2 men and 2 women, we know that we must take a group of 2 men out of the group of 15 men and a group of 2 women out of the group of 20 women. Therefore, we have:

    (15 choose 2) x (20 choose 2)

    (15 choose 2) = 105

    (20 choose 2) = 190

    190*105 = 19950

    Therefore, there are 19950 ways to have a group of 4 with 2 men and 2women.

    If there has to be 1 man and 3 women, we know that we must take a group of 1 man out of the group of 15 men and a group of 3 women out of the group of 20 women. Therefore, we have:

    (15 choose 1) x (20 choose 3)

    (15 choose 1) = 15

    (20 choose 3) = 1140

    15*1140 = 17100

    Therefore, there are 17100 ways to have a group of 4 with 3 women and 1 man.

    We now find the total outcomes of having a group with 4 women.

    We know this is the same as saying (20 choose 4) = 4845

    Therefore, there are 4845 ways to have a group of 4 with 4 women.

    We now add the outcomes of 2 women, 3 women, and 4 women and get the total ways that a committee can have at least 2 women.

    19950 + 17100 + 4845 = 41895 ways that there will be at least 2 women in the committee
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