$\left[\begin{array}{ll}[\text { Classical combinatorics (other) })]\end{array}\right]$

On a $11 \times 11$ grid, 22 cells are marked such that exactly two cells are marked in each row and each column. Two arrangements of marked cells are equivalent if, by swapping any number of columns among themselves and any number of rows among themselves, one arrangement can be transformed into the other. How many non-equivalent arrangements of marked cells exist?
Consider a bipartite graph, where the vertices are the rows and columns of the given board, and a row and a column are connected by an edge if and only if the cell at their intersection is marked. Note that permutations of the rows and columns of the original table correspond to renumberings of the vertices in each part of the resulting graph. This means that equivalent placements correspond to isomorphic graphs, and non-equivalent placements correspond to non-isomorphic graphs. Thus, the number of non-equivalent placements of marked cells is equal to the number of non-isomorphic graphs corresponding to some placements. Let's see what graphs correspond to some placements of marked cells. First, each part of this graph must have 11 vertices (the part of rows and the part of columns). Second, the degree of each vertex is 2, which means the graph decomposes into cycles. Since the graph is bipartite, the lengths of these cycles are even. Moreover, since the graph has no multiple edges, the length of each such cycle is at least four. Isomorphism of two such graphs is equivalent to the coincidence of their sets of cycle lengths. Therefore, the desired number is the number of representations of the number 22 as a sum of even addends, each of which is at least four (sums differing only in the order of addends are considered the same), which is the number of representations of the number 11 as a sum of addends, each of which is at least 2. By listing all these representations, one can verify that there are exactly 14 of them.

## Answer

14 placements.