How many to color the fields of a chessboard?

Question

From Pavle Mladenović's Combinatorics: A Problem-Based Approach

How many ways are there to color the fields of a chessboard $8 \times 8$ using $8$ colors, such that every color appears in each row, and no two adjacent fields in the same column are of the same color?

Since the derangement formula is !$n = \frac{n!}{e}$. Will this just be $\frac{8!}{e}$?

$\frac{n!}{e}$ isn't an integer. It's very close to $!n$, though. — Arthur, Dec 18 '19 at 07:27
Maybe this https://math.stackexchange.com/questions/92752/color-an-n-times-n-square-with-n-colors?rq=1 answers your question? — Math, Dec 18 '19 at 07:37

Arthur · Accepted Answer · 2019-12-18T07:31:54.273

3

You're close, but you're thinking too small.

There are $8!$ ways to colour the first row (and we are already beyond your $!8$ suggestion). Then there are $!8$ ways to colour the second row. And then there are $!8$ ways to colour the third row. And so on.

So the answer is that there are $8!\cdot (!8)^7$ ways to colour the chessboard under these restrictions.

edited Dec 18 '19 at 07:31

answered Dec 18 '19 at 07:30

Arthur

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How many to color the fields of a chessboard?

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