# Arrangement of $4$ pieces on a chess board

Q. What is the total number of ways in which four distinguishable pieces be placed on an $8$X$8$ chess board such that no two pieces are in the same row or column?

This question is remarkably similar to the question in which there are eight pieces, but this case is different, as both cannot be solved by the same method.

If there were eight pieces, the number of ways would have been simply $(8!)^2$. That could be thought as one piece per each row, with the piece in the first row having $8$ possible places, the second having $7$ and so on, giving $$8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot2 \cdot1 = 8!$$ for each row. Multiplying the $8!$ cases for each column, we get $(8!)^2$

I am not able to think of any helpful approach to the four piece case.

How can the four piece case be solved?

• As the answers point out, your answer for eight pieces is not correct. It should be $(8!)^2$ and you can use the same technique to get there. – Ross Millikan Jun 22 '17 at 14:47
• @RossMillikan I did not notice the mistake. Thanks for correcting. – Ananth Kamath Jun 22 '17 at 14:52
• I have added it to the question now. – Ananth Kamath Jun 22 '17 at 14:58

Let's put a piece on one of the $64$ squares on your chessboard. Now you have eliminated one row and one column, so you are left with a $7\cdot7$ chessboard. You can put your piece in one of that $49$ squares. For the third one, as you can imagine, you'll have a $6\cdot6$ square and so on.

So, in total, you have $8^2\cdot7^2\cdot6^2\cdot5^2$ possibilities.

The first piece has $64$ possible locations to choose from, the second has $64-15=49$, the third has $49-13=36$, and the fourth has $36-11=25$

So the total number is $$64\times 49\times 36\times 25 = 2822400$$

You can put piece $1$ on any of the 64 squares. You then come to piece $2$. It must be in a different row and column to piece $1$. How many possible square are there for piece $2$? piece $3$? piece $4$?

• Is the answer $8^2 \cdot 7^2 \cdot 6^2 \cdot 5^2$? – Ananth Kamath Jun 22 '17 at 14:46
• @AnanthKamath Aha! – Lord Shark the Unknown Jun 22 '17 at 14:48
• Aha indeed! Thanks, @Lord Shark the Unknown! – Ananth Kamath Jun 22 '17 at 14:52