1.74 Suppose the network lines drawn on the paper consist of $n$ horizontal lines and $n$ vertical lines. How many different closed broken lines of $2 n$ segments can be drawn along these network lines, such that each broken line passes through each horizontal line and each vertical line?
[Solution] A closed polyline along the network lines must be a combination of horizontal and vertical segments, so it must have $n$ horizontal segments and $n$ vertical segments. It is easy to see that the polyline is uniquely determined by its set of vertices, and the set of vertices is uniquely determined by a permutation of horizontal and vertical segments
$$
\left\{i_{1}, j_{1}, i_{2}, j_{2}, \cdots, i_{n}, j_{n}\right\}
$$

where $\left\{i_{1}, i_{2}, \cdots, i_{n}\right\}$ and $\left\{j_{1}, j_{2}, \cdots, j_{n}\right\}$ are both permutations of $(1,2, \cdots, n)$. In this way, we obtain a mapping from permutations (1) to polylines, which is clearly surjective but not injective. For any $1 \leqslant k \leqslant n$, the permutations
$$
\begin{array}{l}
\left\{i_{k}, j_{k}, \cdots, i_{n}, j_{n}, i_{1}, j_{1}, \cdots, i_{k-1}, j_{k-1}\right\}, \\
\left\{i_{k}, j_{k-1}, i_{k-1}, j_{k-2}, \cdots, j_{1}, i_{1}, j_{n}, i_{n}, \cdots, i_{k+1}, j_{k}\right\}
\end{array}
$$

correspond to the same polyline, so this mapping is a multiple mapping with a factor of $2n$. Since the number of all permutations of the form (1) is $(n!)^{2}$, the number of different closed polylines is $\frac{1}{2} n!\cdot(n-1)!$.