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?

  • $\begingroup$ 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? $\endgroup$
    – bof
    Oct 7 '20 at 9:47
  • 1
    $\begingroup$ 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? $\endgroup$
    – bof
    Oct 7 '20 at 9:49
  • $\begingroup$ 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. $\endgroup$ Oct 7 '20 at 9:58
  • $\begingroup$ 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. $\endgroup$
    – Math Lover
    Oct 7 '20 at 9:59
  • $\begingroup$ @dudeguyguerilla pls see my answer which I have edited to using your method. I hope that explains the undercounting. $\endgroup$
    – Math Lover
    Oct 7 '20 at 10:28


Adding your method and completing it which works too -

enter image description here

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.


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