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There are 8 men and 7 women from which a group of 4 with at least 2 men must be selected. Find how many possible groups there are.

There are 2 ways which both make sense to me but only one method is right.

Correct method:

(No. of groups with 2 men, 2 women) + (No. of groups with 3 men, 1 woman) + (No. of groups with 4 men)

$_8C_2 \times _7C_2 + _8C_3 \times _7C_1 + _8C_4=1050$

Incorrect method:

No. of ways of choosing 2 men from 8 = $_8C_2$

This leaves the remaining 2 spots unrestricted so 2 people can be chosen from the remaining 13 people.

$_8C_2 \times _{13}C_2 = 2184$

Any idea why this second method is incorrect?

Eleven-Eleven
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rattatttta
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  • The best way i can explain is this: by allowing the second choice to contain the remaining 6 men and 7 women, you are essentially rechoosing the men for the remaining two slots, when in the correct situation, you first choose the two men and then the two women, plus the possibility that there are three men, ... hope that helps. – Eleven-Eleven Nov 04 '13 at 02:30

1 Answers1

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The incorrect method counts some possible groups more than once. One possible group (using tradition male/female names to indicate sex) is {Juan, Mary, Patrick, Luke}. The second method will count it three times: once where Juan and Luke are the 2 men chosen first, once with Juan and Patrick, and once with Patrick and Luke.

Steve Kass
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