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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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I now regret that I got my degree in math rather than mayhem. :) – Austin Mohr May 3 '12 at 0:45
I suspect you've left out an important part of the first problem. Surely you mean that not only are no two "towers" in the same row, but also no two in the same column. Without the latter restriction the count would be $8^8$ rather than $8!$. – hardmath May 3 '12 at 1:59
@hardmath: I added that information to the question. – spohreis May 3 '12 at 21:54
up vote 5 down vote accepted

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.

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