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On a quiz, I asked the following question.

A group of 15 students contains seven boys and eight girls. In how many ways can a committee of 5 be selected if it must contain at least one girl?

I know the answer is $\binom{15}{5}-\binom{7}{5}=2982$ committees.

I cannot figure how to explain the error in the following approach.

The committee needs a girl, which has $\binom{8}{1}=8$ ways. Once a girl is selected, I don't care which 4 of the 14 persons are selected, which has $\binom{14}{4}=1001$ ways of happening. Thus, there are $8\cdot14=8008$ committees possible.

Can someone help me explain the error in logic here?

Tim Thayer
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  • The problem is that you are double-counting some possibilities. You are considering the choice of the first girl as dependent on the order that it happened. But it may happen that you choose a different girl at first, and then choose the same remaining set of people afterwards. – Tob Ernack Mar 05 '18 at 02:43

2 Answers2

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Your answer overcounts. Let us label all the 15 people: $B_1,\ldots,B_7$ are the boys and $G_1,\ldots,G_8$ are the girls.

What you are doing with $\binom{8}{1}\binom{14}{4}$ is picking a girl and then forming a committee of four from the remaining 14. This double counts cases like the committees $G_1,B_1,B_2,B_3,G_2$ and $G_2,B_1,B_2,B_3,G_1$ (they both should be the same, but picking the girl induvidually gives the duplicate).

tejas
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It leads to overcounting. Say you first select Sara. Then from remaining you select Liz and Rachel and some other boys.

This will be same as selecting Liz first and then from remaining selecting Rachel, Sara and the same boys.

King Tut
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