5. A party has prepared $2 n$ gifts, which are evenly divided into two strings and publicly hung on the wall. Each winner can (and can only) choose one gift from the bottom of either string. If there are $2 n$ winners selecting these $2 n$ gifts, then there are $\qquad$ different ways of selection.
5. $\mathrm{C}_{2 n}^{n}$.
Suppose the two strings of gifts from bottom to top are:
$$
a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n} \text {. }
$$

Consider the $2n$ winners as $2n$ positions. Choosing $n$ positions from $2n$ positions, there are $C_{2 n}^{n}$ ways to choose. Arranging $a_{1}, a_{2}, \cdots, a_{n}$ in the chosen positions in increasing order of their indices has 1 way; arranging the remaining $n$ positions with $b_{1}, b_{2}, \cdots, b_{n}$ in increasing order of their indices also has 1 way. Therefore, the $2n$ winners selecting these $2n$ gifts have a total of $\mathrm{C}_{2 n}^{n}$ different ways.