2 white and 2 black rooks on chessboard I realize this is a fairly basic combinatorial problem, but I'm still confused. 
In how many was can 2 black and 2 white rooks be placed on a standard chessboard such that white and black don't attack each other. 
The way I saw it first, there are 2 cases:
Case 1. Both white are on the same row/column. In this case there are $
\frac{64 \times 15}{2} \times \binom{49}{2}$ ways of allocating these pieces.
Case 2. Both white are on different rows and columns. In this case there are $\frac{64 \times 49}{2} \times \binom{34}{2}$
Hence the solution is the sum of Case 1 and Case 2. After some deliberation, I realized that this is probably wrong because the order in which the pieces are placed on the board matter, e.g. WBWB is different to WWBB. 
So I came up with a different idea, to use inclusion-exclusion theorem.
First, we find 4 squares: $\binom{64}{4}$ for all pieces, then we have 4 cases: 1w/1b attack each other, 2w/1b attack each other, 1b/2w attack each other and 2w/2b attack each other, but as I started analyzing them, I got stuck (e.g. case 1 depends on the distance between rooks), and not sure this is the correct approach either. 
Any suggestions? 
 A: I'm copying your own solution:
There are 2 cases:
Case 1. Both white rooks are on the same row/column. In this case there are $
\frac{64 \cdot 14}{2} \cdot \binom{42}{2}$ ways of allocating the four pieces.
Proof. The first white rook may be placed on $64$ squares. Then there are $14$ squares for the second white rook. Since the two white rooks are indistinguishable we have to divide $64\cdot14$ by $2$. Now $2$ rows and $1$ column (or vice versa) are forbidden for the black rooks. There remain $6\cdot7$ allowed squares, and we can freely choose $2$ of them.
Case 2. Both white rooks are on different rows and columns. In this case there are $\frac{64 \cdot 49}{2} \cdot \binom{36}{2}$ ways of allocating the four pieces.
Proof. Essentially the same as in case 1.
A: Here is an alternate, slightly more complicated, solution:
$\textbf{1)}$ If none of the rooks attack each other, there are $\displaystyle\binom{8}{4}\binom{8}{4}\big(4!\big)\binom{4}{2}=\color{blue}{705,600}$ possibilities:
$\hspace{.4 in}\binom{8}{4}$ ways to choose their columns, $\binom{8}{4}$ ways to choose their rows, 
$\hspace{.4 in}4!$ ways to place them in this $4\times4$ array, and $\binom{4}{2}$ ways to choose the rooks which are black
$\textbf{2)}$ If only 1 pair of rooks attack each other, there are $\displaystyle16\binom{8}{2}\binom{7}{2}\binom{6}{2}\big(2!\big)\binom{2}{1}=\color{blue}{564,480}$ choices:
$\hspace{.4 in}$ 16 ways to choose the row (or column) for this pair, $\binom{8}{2}$ ways to choose their columns (or rows), 
$\hspace{.4 in}\binom{7}{2}$ ways to choose the rows (or columns) for the other two rooks, $\binom{6}{2}$ ways to choose their columns 
$\hspace{.4 in}$(or rows), $2!$ ways to place them in this $2\times2$ array, and $\binom{2}{1}$ ways to choose the color for the first pair
$\textbf{3)}$ If two pairs of rooks attack each other, there are $\displaystyle16\binom{8}{2}\left[7\binom{6}{2}+6\binom{7}{2}\right]=\color{blue}{103,488}$ possibilities:
$\hspace{.4 in}$ 16 ways to choose the row (or column) for the black pair, $\binom{8}{2}$ ways to choose their columns (or rows),
$\hspace{.4 in}$ 7 ways to choose the row (or 6 ways to choose the column) for the white pair, and then
$\hspace{.4 in}\binom{6}{2}$ ways to choose their columns (or $\binom{7}{2}$ ways to choose their rows).
This gives a total of $\color{blue}{1,373,568}$ possibilities.
