# How many ways are there to position two black rooks and two white rooks on an 8X8 chessboard

## How many ways are there to position two black rooks and two white rooks on an 8X8 chessboard so that no two pieces of different colors share a row or a column.

I have trouble understanding the solution which is: There are $64*49/2$ ways two position two black rooks so that they don't share a row or a column and $64*14/2$ ways to position them in the same row or in the same column. In the 1rst case there are six rows and six columns which remain available for the white rooks and so there are $36*35/2$ ways to position them....

Why do we multiply by 49?

• Suppose one black rook has been placed on the chessboard, say on the lower left corner square. How many ways are there to place the other black rook, so that it's not on the same row or column as the first one? – bof Nov 20 '14 at 23:53

The number of ways to place the white rooks depends on whether the black rooks share a row/column or not.

If they do not, which is the "first case" the solution talks about, you'd want to count the number of ways that two black rooks can be placed so that they don't share a row or column.

The first rook has $64$ possibilities. Once you've placed it, you eliminate $15$ possibilities for the second rook's position ($7$ in one row and one column, plus the square that the first rook is on, so $7 \cdot 2 + 1 = 15$).

Since $15$ possibilities are eliminated, there are $64-15=49$ possible placements for the second rook.

That's where the $49$ comes from.

• Totally missed that. Thank you! – Basil M. Nov 21 '14 at 0:06

There are $64$ ways to position the first black rook.

There are $49$ ways to position the second black rook such that it doesn't share the same column as the first (case 1), and $14$ ways to position it such that it does (case 2).

For case 1, there are $36$ ways to place the first white rook such that it doesn't share a row or column with either black rook, and $35$ ways to position the second so it does the same.

For case 2, there are $42$ ways to place the first white rook and $41$ ways to place the second.

The same coloured rooks are interchangeable, so divide by $4$.

That is: $\dfrac{64\times(49\times 36\times 35+14\times 42\times 41)}{2\times 2}$