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How many ways are there to choose a committee from $n$ boys and $n$ girls such that there are strictly more boys than girls?

You can express your answer in exponents and factorials.

  • What have you tried? – David G. Stork Apr 14 '20 at 00:22
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    Hint: there are as many of this sort as there are committees with strictly more girls than boys. – lulu Apr 14 '20 at 00:24
  • I just know that 0g and 1b is n choose 1, 0g and 2b is n choose 2, ..., 0g and nb is n choose n, so they sum up to 2^n –  Apr 14 '20 at 00:25
  • Your most recent comment talks only about choosing zero girls. Also, the sum you describe $\binom{n}{1}+\binom{n}{2}+\dots+\binom{n}{n}$ is not quite equal to $2^n$ since you didn't include $\binom{n}{0}$ in that sum. It in fact added to $2^n-1$ instead. Now... you could continue that argument and now consider the cases where there were one girl and either $2,3,4,\dots,n$ boys, then where there were two girls and some number of boys, etc... – JMoravitz Apr 14 '20 at 00:35

2 Answers2

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Let's assume that a "committee" just means a subset of the $2n$ kids. There is some ambiguity there since it isn't clear if the empty set is a committee or not, but that won't matter here as the condition isn't satisfied. We'll allow the empty set to be a committee...it's easy to see that the final result won't change if that case is excluded.

There are $2^{2n}$ possible committees. These come in three types: more boys, more girls, same number of boys and girls. Clearly the first two types have the same number, call it $F(n)$. The number of committees with the same number of boys and girls is $$\sum_{i=0}^n\binom ni^2=\binom {2n}n$$

(for a proof of this equality, see this question)

Thus we have $$2F(n)+\binom {2n}n=2^{2n}\implies \boxed {F(n)=\frac 12\times \left(2^{2n}-\binom {2n}n\right)}$$

lulu
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Let $b$ be the number of boys, and $g$ be the number of girls. There are $n$ boys to pick from and $n$ girls to pick from:

$$\sum\limits_{b=1}^n \sum\limits_{g=0}^{b-1} {n \choose b} {n \choose g}$$

The cleverer way to approach it is find the number of ways you can get exactly the same number of boys and girls:

$$K= \sum\limits_{k=0}^n \left( {n \choose k} \right)^2$$

Then take the total number of ways you can choose any number of boys and any number of girls, call it $R$. Subtract the above ($K$) from this number, i.e., $R-K$. In half of these the boys outnumber the girls; in the other half the girls outnumber the boys.

Can you do the arithmetic?

David G. Stork
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