Given $N,M$ with $1 \le M \le 6$ and $1\le N \le 36$. In how many ways we can place $N$ knights (mutually non-attacking) on an $M \times M$ chessboard?
For example:
$M = 2, N = 2$, ans $= 6$
$M = 3, N = 2$, ans $ = 28$
$M = 6, N = 15$, ans $= 2560$
It is possible to solve this using graph theory. However, I am (more) interested in a combinatorial approach (A closed form solution).
Best Answer
The complete answer to this question is contained in the Knights polynomials. By definition, the coefficient of $q^k$ in the $m\times n$ knights polynomial, $K_{m,n}(q)$, is the number of ways that $k$ non-attacking knights can be placed on an $m\times n$ board. Of course, it's a tautology to say that this solves the problem, but there is an algorithm to compute them. Using this algorithm, the first six $m\times n$ knights polynomials (per your request) are
\begin{align} K_{1,1}(q) &= q+1 \\ K_{2,2}(q) &= q^4+4 q^3+6 q^2+4 q+1 \\ K_{3,3}(q) &= 2 q^5+18 q^4+36 q^3+28 q^2+9 q+1\\ K_{4,4}(q) &= 6 q^8+48 q^7+170 q^6+340 q^5+412 q^4+276 q^3+96 q^2+16 q+1\\ K_{5,5}(q) &= q^{13}+15 q^{12}+158 q^{11}+994 q^{10}+3679 q^9+8526 q^8+12996 q^7+13384 q^6 \\ &+ \; 9386 q^5+4436 q^4+1360 q^3+252 q^2+25 q+1\\ K_{6,6}(q) &= 2 q^{18}+40 q^{17}+393 q^{16}+2560 q^{15}+12488 q^{14}+47684 q^{13}+141969 q^{12} \\ &+ \; 323476 q^{11}+556274 q^{10}+716804 q^9+688946 q^8+491140 q^7+257318 q^6 \\ &+ \; 97580 q^5+26133 q^4+4752 q^3+550 q^2+36 q+1 \end{align}
Note that these polynomials agree with the results that you provide. For example, the coefficient of $q^{15}$ in $K_{6,6}$ is 2560, corresponding to your statement that $M=6,N=15 \Rightarrow \text{ans} =2560$. The coefficients of $q^2$ also agree with this OEIS sequence.
The algorithm to compute these is very similar to the obviously related Kings problem. Neil Calkin and his REU students provide a very thorough analysis of that problem in two papers in Congressus Numerantium. Shaina Race has a publicly accessible presentation of their results here. I published a paper presenting an implementation of the algorithm in Mathematica V5. The technique for counting knights is a bit more complicated but uses very similar ideas. Unfortunately, I've been able to compute the $7\times 7$ knights polynomial but my computer chokes on $8\times 8$, the size of an actual chess board.