# Chessboard pawns arrangement clarification

I have a 8 X 8 chessboard, and 8 identical pawns. These pawns are arranged at random. What is the probability that the pawns are arranged in such a way that each row and column have only one pawn?

My solution:

Since these are 64 squares, and 8 pawns, we can choose any of the 8 squares from 64 squares in $${64 \choose 8}$$ ways. These are the exhaustive number of cases.

Now the favorable number of cases, can be found by thinking that each row has only one favorable position to place the pawn (the diagonal position). So the 8 pawns can be placed on diagonal in $$8 . 7 . 6 . 5 . 4 . 3 . 2 . 1$$ or $$8!$$ ways. But since we have 2 diagonals, we have to multiply the $$8!$$ with 2, giving the required probability as

$$\frac{8! . 2 }{64 \choose 8} = 0.000018$$ But the answer is supposed to be half of this answer 0.000009, which should have obtained by considering just one diagonal. Could you point out my mistake, and with possible explanation of why we should not consider the 2 diagonals.

Sekhar

The pawns need not be on the diagonals. And you have counted the number of way to place the pawns on the diagonals incorrectly, there are only $2$ ways. For remember that $\binom{64}{8}$ counts the number of ways to choose the set of positions of the pawns. So in counting favourables, we must also count sets of positions only. The top left to bottom right diagonal is just $1$ set of positions.
The pawn in the first row can be placed in $8$ ways. For each such way, the pawn in the second row can be placed in $7$ ways, and so on, for a total of $8!$ favourables.
• There are two diagonals of length $8$. The probability the pawns are on one or the other is $\frac{2}{\binom{64}{8}}$. Much less than your proposed number. If you mean one specific diagonal, say top left to bottom right, replace the $2$ by $1$. Commented Aug 27, 2015 at 4:28