Answer:
[tex]\frac{(n!)^{2} }{k!((n-k)!)^2}[/tex]
Step-by-step explanation:
First, we select the columns from our n columns where we will place our k rooks. We use a combinatoric for this:
[tex]\left[\begin{array}{ccc}n\\k\end{array}\right][/tex]
Using the same line of reasoning, we choose the rows to place our k rooks from our n rows:
[tex]\left[\begin{array}{ccc}n\\k\end{array}\right][/tex]
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 = [tex]\left[\begin{array}{ccc}n\\k\end{array}\right][/tex] * [tex]\left[\begin{array}{ccc}n\\k\end{array}\right][/tex] * k! = [tex]\frac{n!}{k!(n-k)!} *\frac{n!}{k!(n-k)!}*k! =\frac{(n!)^{2} }{k!((n-k)!)^2}[/tex]
