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?

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:

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

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 = * * k! =

+25 = x

step-by-step explanation:

here x is the hypotenuse of a right triangle whose legs are 7 and 24.   by the pythagorean theorem, 7² + 24² must equal x²:

x² = 49 + 576 = 625.   thus, x = +√625 = +25 = x

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