Consider an $8 \times 8$ chessboard.

How many ways are there to place 6 rooks such that no two rooks are on the same row or column?
Let $n$ be the size of the chessboard and $k$ the number of rooks.

First, we choose the $k$ columns for the rooks. There are $\binom{n}{k}$ possibilities. We then choose the rows for the rooks one by one, which gives $n \times(n-1) \times \cdot \times(n-k+1)$ possibilities.

In total, we have

$$
\binom{n}{k} \times n \times(n-1) \times \cdot \times(n-k+1)=\binom{n}{k} \frac{n!}{(n-k)!}
$$

which can be rewritten as

$$
\binom{n}{k}^{2} k!
$$

With the numerical values from the problem, we get

$$
\binom{8}{6}^{2} 6!=(28)^{2} \times 720=564480
$$

## 6 Inequalities (Angela and Émile)

## Arithmetic-Geometric Inequality

Theorem 1 (Arithmetic-Geometric Inequality).

Let $a, b$ be real numbers. Then we have

$$
\frac{a^{2}+b^{2}}{2} \geq a b
$$

with equality if and only if $a=b$.

## Remark 2.

If $a$ and $b$ are strictly positive real numbers, we have

$$
\frac{a+b}{2} \geq \sqrt{a b}
$$

with equality if and only if $a=b$.