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I am self-studying Discrete Mathematics and I want to solve the following question.

How many ways there are to put eight equal towers on a chessboard such that there are no two equal towers in the same row or in the same column? And if the towers are distinct-looking?

I solved the first part, and the answer is $8!$, but I was not able to solve the second part. It says that the answer is $(8!)^{2}.$ Could you please help me to solve this?

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    @hardmath: I added that information to the question.2012-05-03

2 Answers 2

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Imagine you make a solution to the second problem by taking a solution to the first problem and replacing the eight equal towers with eight distinct-looking towers.

For each possible picture in the answer to the first question, you may put your eight distinct-looking towers in the same spots that are being occupied by equal towers in 8! different ways (because there are 8! different ways to order the eight distinct-looking towers, and for each possible order, you can put tower $n$ with respect to that order in row $n$).

So the answer is 8! times the answer to the first question.

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The first one is easy.

If the towers are different, then there are 8! ways of choosing the 8 positions where they must sit, and for each such choice there are 8! ways of putting 8 different objects on those possitions.