$2 n$ men and $2 n$ women can be paired into $n$ mixed (quartet) groups in how many ways? (Tennis game!)





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The first quartet can be chosen separately 2 men and 2 women from $2 n$ individuals in $\binom{2 n}{2}$ ways; the 4 members of the group can be paired in $\binom{2 n}{2}\binom{2 n}{2}=\binom{2 n}{n}^{2}$ ways. However, within a quartet, they can stand in two ways. (I.e., in the arrangement of $a_{2} b_{2}$ and $a_{1} b_{1}$ or $a_{1} b_{2}$ and $a_{2} b_{1}$). According to this, the number of mixed pairs is: $2\binom{2 n}{2}^{2}$. i. t.

The next quartet can be chosen from $2 n-2$ men and $2 n-2$ women; the number of possibilities is: $2\binom{2 n-2}{2}^{2}$, etc.

Accordingly, the number of arrangements of $n$ mixed-pair groups is:

$$
\begin{gathered}
2^{n}\left[\binom{2 n}{2}\binom{2 n-2}{2}\binom{2 n-4}{2} \ldots\binom{4}{2}\binom{2}{2}\right]^{2}= \\
=\frac{2^{n}}{2^{2 n}}[2 n(2 n-1)(2 n-2)(2 n-3)(2 n-4)(2 n-5) \ldots 4 \cdot 3 \cdot 2 \cdot 1]^{2}=\frac{(2 n!)^{2}}{2^{n}}
\end{gathered}
$$

However, in this number, a certain $n$ mixed-pair group appears in every possible order, i.e., $n!$ times. Without regard to order, the number of arrangements is

$$
\frac{(2 n!)^{2}}{n!2^{n}}
$$

Julia Sándor (Ferenc Kölcsey g. VIII. o. Bp. VI.).