# Ways to place 3 chess pieces so that none are in the same column or row

How many ways can we place $$3$$ chess pieces so that none are in the same column or row? Chess pieces are distinguishable, so we can imagine them as one pawn, one knight, one rook. A chess board is a $$8\times 8$$ grid.

So the correct answer is $$8^27^26^2$$ because we have $$8$$ choices for the column and $$8$$ for the row to place the first piece, and $$7$$ choices for the second piece, and so on.

However, when I attempted the question, I thought that we could do $$(8 C 3)\cdot (8 C 3)$$, which first counts the number of ways to choose $$3$$ different columns and then $$3$$ different rows. However that answer seems to undercount. So why is it wrong and why is it undercounting?

• Suppose you had to place $8$ chess pieces instead of $3$. What number does your method give? Is it an undercount? Do you see why it's an undercount?
– bof
Oct 7 '20 at 9:47
• Or maybe this is easier. How many ways can you place a pawn, a knight, and a rook on a $3\times3$ chessboard, so that no two are in the same column or row?
– bof
Oct 7 '20 at 9:49
• Yeah my method would give 1, and I think I see where it's undercounting. I'm referencing a chessboard here, and suppose we are choosing 2 pieces instead of 3 (to make it easier). If we want positions A1 and H8, we choose columns A and H using $8C2$, and we then choose rows 1 and 8 again using $8C2$. However, this also includes A8 and H1 as one way and as a result undercounts. Oct 7 '20 at 9:58
• When you choose 3 different columns and 3 different rows, now you need to choose different squares from them to place your pieces. You are not counting them. Oct 7 '20 at 9:59
• @dudeguyguerilla pls see my answer which I have edited to using your method. I hope that explains the undercounting. Oct 7 '20 at 10:28

EDIT:

You first choose $$3$$ rows and $$3$$ columns so number of ways = $$(^8C_3)^2$$.

This gives you $$9$$ squares. Please see one of the cases in the picture. Now you can place the first piece on any of the $$9$$ squares. That leaves $$4$$ choices for the next piece. The last one has just one places to go to.

So total number of ways = $$(^8C_3)^2 \times 9 \times 4 = 112896$$.

May be an easier way to look at it is

Number of ways $$= 8 \times 8 + 7 \times 7 + 6 \times 6$$. Explanation -

There are $$64$$ squares on a chessboard. The first one can be placed anywhere out of $$64$$ squares - (say, Col1 and Row1). Once that is done, the second one has to avoid squares in col1 and row1. That leaves $$64-15 = 49 = 7 \times 7$$ squares for the second.

Now think of the third one - $$2$$ squares to avoid will be common between first and second. So add another $$13$$ squares to avoid. That leaves $$36 = 6 \times 6$$ squares.