Stones on a chessboard
Task number: 3492
In how many ways can we place eight stones on a \(4 \times 4\) chessboard such that there are four stones in a single row or a single column?
Solution
Let \(A_i\) denote the positions where the entire \(i\)-th row is occupied for \(i\in\{1,…,4\}\), and where the entire \((i-4)\)-th column is occupied for \(i\in\{5,…,8\}\).
We have \(|A_i|=\binom{12}{4}=495\). For the intersections \(A_i \cap A_j\) we must distinguish three situations. When we occupy two rows, i.e. \(i,j\in\{1,…,4\}\), the intersection has only a single element, because there are no stones left. Similar reasoning applies for two columns. When we occupy one row and one column, a single stone remains and we can place it on any of the remaining nine squares.
The intersections of three sets are empty, since we don‘t have enough stones to occupy three rows or columns.
Altogether \(|\bigcup\limits_{i=1}^8 A_i|=8\binom{12}{4}-2\binom{4}{2}-4^2{\cdot} 9=3{,}804\).
Answer
The number of possible arrangements is 3,804.
