Chessboard math

There are $\dbinom{10}{6}$ ways to choose the columns in which rooks will be placed (all columns have to be distinct). Then after fixing the columns , there are $^{10}P_6$ ways to arrange the rooks in columns so that they wont attack anyone. Thus there are $\dfrac{10!}{4! \times 6!} \times \dfrac{10!}{4!}$ ways i.e. $31752000$ ways hence product of digits is $\boxed{210}$

The correct answer is 3072, not 210.

You can place the first rook on any of the $10^2$ squares. Once you place the first rook, there are $9^2$ ways to place the second rook, then $8^2$ ways to place the third rook, and so on, until there are $5^2$ ways to place the sixth rook.

Then $10^2 \cdot 9^2 \cdot 8^2 \cdot 7^2 \cdot 6^2 \cdot 5^2 = 22861440000,$ and $2 \cdot 2 \cdot 8 \cdot 6 \cdot 1 \cdot 4 \cdot 4 = 3072$ .

Since we require 6 rows and 6 columns to place the took, we can choose it by c(10,6)^2. Now arrange them in a single 6×6 square. The no. Of ways to place 6 identical rooks in non attacking positions here is 6!. Thus total no. Of placements is equal to c(10,6)^2*6! Which yields 31752000 as the answer.

We can solve this with the 10th rook polynomial (in the $n \times n$ case). To so, we need the coefficient in the $x^6$ term, as that will be our answer. Due to the identity relating it to Laguerre polynomials $R_{10}(x) = 10! \times x^{10}\times L_{10}(-x^{-1})$ therefore the coefficient for the $x^6$ term in $R_{10}$ will be the same as the coefficient in the $x^4$ term of $L_{10}(x)$ after factoring out the factorial. WolframAlpha to the rescue: the coefficient in front of $x^4$ is $31752000.$ $3\times 7\times 5\times 2 = \boxed{210}$