Given an checkerboard with alternating white and black squares, how many ways are there to choose four black squares and four white squares so that no two of the eight chosen squares are in the same row or column?
Problem 1296
Official solution
Solution:
Number both the rows and the columns from to , and say that black squares are the ones where the rows and columns have the same parity. We will use, e.g., "even rows" to refer to rows . Choosing squares all in different rows and columns is equivalent to matching rows to columns.
For each of the rows, we first decide whether they will be matched with a column of the same parity as itself (resulting in a black square) or with one of a different parity (resulting in a white square). Since we want to choose squares of each color, the rows matched to same-parity columns must contain even rows and odd rows. There are ways to choose odd rows and even rows to match with same-parity columns.
After choosing the above, we have fixed which rows should be matched with odd columns (while the other should be matched with even columns). Then there are ways to assign the columns to the rows, so the answer is .