Question

In chess, a rook attacks all cells on the row and column that it is in. Let k ≤ n. How many ways are there to place k rooks on an n × n chess board in such a way that no two rooks attack one another?

Answer 1

`((n!)^(2) )/(k!((n-k)!)^2)`

Explanation:

First, we select the columns from our n columns where we will place our k rooks. We use a combinatoric for this:

`\left[\begin{array}{ccc}n\\k\end{array}\right]`

Using the same line of reasoning, we choose the rows to place our k rooks from our n rows:

`\left[\begin{array}{ccc}n\\k\end{array}\right]`

Now, we need to form cells using the colums and rows we selected. For example, imagine a 3x3 board, where you will place 2 rooks. We select the first 2 columns and the firsts 2 rows to place the rooks. The cells where the rooks are gonna be placed are formed by pairing columns and rows, like this:

- First choose the Column 1. You have 2 rows available for pairing: Row 1 and Row.

- Once you formed the first cell, you will have left 1 column and 1 row left to place the remaining Rook.

Extrapolating this logic for k Rooks, you will notice that the number of possible ways to distribute the rooks will always be k!.

Finally, the total amount of ways there are to place k rooks on an n × n chess board in such a way that no two rooks attack one another will be the multiplication of the 3 expressions we develop so far:

Total =
`\left[\begin{array}{ccc}n\\k\end{array}\right]` *
`\left[\begin{array}{ccc}n\\k\end{array}\right]` * k! =
`(n!)/(k!(n-k)!) *(n!)/(k!(n-k)!)*k! =((n!)^(2) )/(k!((n-k)!)^2)`