# How many possible setups on the chessboard?

Jason has $5$ indistinguishable white pawns and $4$ indistinguishable black pawns that he wishes to place on a $3 \times 3$ chessboard (where each square of the chessboard is distinct) such that there is exactly one pawn per square. He describes a position as drawish if no row or column has pawns of all the same color. How many possible setups are drawish?

Attempt:

We have that since $A \cup B = A + B - (A \cap B)$, that the number of ways to have at least one row or column black is $6 \binom{6}{2}-9$. Similarly the number of ways to have at least one row white or one column white is $6 \binom{6}{1}$ (choose a row or column to be white then we pick one other white tile from the other $6$). Therefore, using $A \cup B = A + B - (A \cap B)$ we can calculate the number having at least one row or column black or at least one row or column white. This is equal to $A^* \cup B^* = 6 \binom{6}{2} - 9 +6 \binom{6}{1}-6*2*3$ (the $6*2*3$ comes from the fact of choosing a black row or column then a white then choosing $2$ of the remaining $3$ tiles to be black. Thus, the answer should be $\binom{9}{4}-A^*\cup B^* = 126-(6 \binom{6}{2} - 9 +6 \binom{6}{1}-6*2*3) = 45$.

Issue:

This answer I provided was not one of the answer choices. It is possible that there was an error with this question and I was right, though. The answer choices were as follows: (A) $36$ (B) $47$ (C) $69$ $(D)$ $75$ (E) $93$.

Your calculation of the number of ways to have a black row or column is fine. You have $6$ rows to choose as the full one, $6 \choose 2$ ways to place the other pawns, and $9$ cases you have counted twice because you have both a full row and full column. If you don't have a black row or column, you also don't have a white row or column because there will be a black pawn to block it. The number of ways to not have a monochromatic line is therefore ${9 \choose 5}-6{6 \choose 2}+9=45$, agreeing with your calculation.