1.49 On a grid paper of size $11 \times 11$, mark 22 cells so that exactly two cells are marked in each row and each column. If one arrangement of the marked cells can be transformed into another arrangement by swapping rows with rows or columns with columns, then these two arrangements are considered equivalent. How many distinct non-equivalent arrangements of the marked cells are there?
[Solution] Consider the centers of 22 marked squares, and connect line segments between any two points in the same row and any two points in the same column. Thus, we obtain a bipartite graph with 22 vertices, each of degree 2. By a theorem in graph theory, this graph must consist of one or several disjoint cycles. Since the edges of the cycles in the graph are either horizontal or vertical, each cycle has an even number of edges, and if a marked point is on a cycle, the marked points in the same row or column are also on the same cycle. Let the graph be divided into $k$ cycles, with the number of point pairs on each cycle being $n_{1}, n_{2}, \cdots, n_{k}$, then $n_{1}+n_{2}+\cdots+n_{k}=11$.

Note that when exchanging columns with columns or rows with rows, each cycle remains a cycle. Moreover, any two cycles with the same number of vertices can be transformed into each other through exchanges of rows or columns. Therefore, two arrangements are not equivalent if and only if the sets of vertex pair numbers $\left\{n_{1}, n_{2}, \cdots, n_{k}\right\}$ are different. And these different sets correspond to the different ways of partitioning 11 into the sum of integers no less than 2. Since there are 14 such partitions: $11,9+2,8+$
$$
\begin{array}{l}
3,7+4,6+5,7+2+2,6+3+2,5+4+2,5+ \\
3+3,4+4+3,5+2+2+2,4+3+2+2,3+ \\
3+3+2 \text { and } 3+2+2+2+2 \text {, it follows that the } 22 \text { marked points }
\end{array}
$$

have 14 non-equivalent arrangements.