In how many ways we can choose $k$ squares on $n \times n$ chessboard, given certain restrictions?

Question

In how many ways we can choose $k$ squares on $n \times n$ chessboard, so that none of these squares are on the same column nor the same row?

Sorry, but the question is the same as in the title. Here are my poor attempts to the problem.

1. First we can choose one square from $n$ rows and $n$ columns so there are $n^2$ ways to do it.

2. Now we can choose one square from $n-1$ column and rows. And so on...

Please help me

Answer 1

We can choose the rows that are used in $\binom nk$ ways, the columns that are used in another $\binom nk$ ways, and then choose which column get assigned to which row in $k!$ ways, for a total of $k!\binom nk^2$ arrangements.

Answer 2

We can go row by row. First, there are $\binom{n}{k}$ ways to choose which rows are in use. Then, there are $n$ ways to make the choice on the top row in use. Then there are $n-1$ ways to make the choice on the next row. In general the $i^{th}$ row will have $n+1-i$ choices, so in total there will be $$ \binom{n}{k}\cdot (n\cdot(n-1)\cdots (n-k+1))=\binom{n}{k}\cdot\frac{n!}{(n-k)!}=\frac{(n!)^2}{k!(n-k)!^2} $$ choices.