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A group of $2N$ boys and $2N$ girls is randomly divided into two equal groups. What is the probability that each group has the same number of boys and girls?

I had a few ideas on this question. Since both groups must have the same number of boys and girls, i counted the numer of ways to form up pairs of boys and girls. Then, i counted the number of ways to divided those pairs into two groups, and since this number of pairs is even, those new groups would be equal in number. Dividing by the number of ways to form up two distinct groups with everybody, this would give me the probability.

Is this right?

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The probability is (assuming all "group divisions" are equally likely) $\frac{M}{K}$, where $M$ is the number of "divisions" in which there are $N$ boys and $N$ girls in each group, and $K$ is the number of "divisions" in total.

$K={{4N}\choose{2N}}$, because this binomial coefficient represents the number of ways in which we can choose $2N$ elements from a set with $4N$ elements, where all elements are different (but we don't care if they're boys or girls here, since there can be an arbitrary number of any in the "first" group). The second group is uniquely determined by the remaining $2N$ people.

$M={{2N}\choose{N}}^2$, because, in order to choose our "favorable" group, i.e. the one that satisfies the condition, we need to select $N$ boys from a group of $2N$ boys, and the same with girls.

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A much simpler approach is to note that one group has $2N$ children. How many ways to select $2N$ children out of $4N?$ Then among the groups of $2N$ children, how many have $N$ boys in them?

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Hint

$$\frac{C (N,2N).C (N,2N)}{C (2N,4N)}$$

where $C (n,p) $ is the binomial coefficient = number of n-elements sets in a p-elements set.

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