In how many different ways can six kings be placed on a $6\times 6$ chessboard so that no one attacks the others?
If the problem was asked for a $3 \times 3$ board and $3$ kings, then the answer would be 8.
I tried to find all the combinations at least two kings are attacking each other. Placed two kings adjacent vertically, horizontally, diagonally and counted the number of possibilities for the remaining $4$ kings. But there seems to be too many overlaps, meaning I counted the same situation more than once. And I couldn't figure out a way to subtract all the situations happening more than once.
Best Answer
If haven't done the calculations, but can tell you how they can be done for the general problem of placing $k$ kings on a $p\times q$ board.
Let $I=\{1,\ldots,p\}$ represent a column of the board. For simplicity, assume that $p\le q$. Let $\cal{C}$ be the set of subsets of $I$ so that no pair $x,x+1\in C$ for any $C\in\cal{C}$: these sets represent permitted placements of kings in a column.
Now we define a matrix $A$ with elements $A_{ij}$ for $i,j\in\cal{C}$: you may enumerate the sets in $\cal{C}$ any way you like. We define the elements $$ A_{ij}=\left\{\begin{array}{l l} 0&\text{if there are $x\in i$, $y\in j$ with $|x-y|\le1$}\\ x^{|j|}&\text{otherwise, where $|j|$ is the number of elements in $j$} \end{array}\right. $$ so $A_{ij}$ represents the addition of a column with kings in the $j$ positions after having had a column with kings in the $i$ positions.
If we let $e=\{\}\in\cal{C}$ represent the initial state, i.e. no kings to the left of the board, and successively add $q$ columns, we get $$ e\cdot A^q\cdot 1=\sum_{k} a_kx^k $$ where the $1$ represents the vector $(1,\ldots,1)$ and $a_k$ is the number of ways to place $k$ kings on the $p\times q$ board.
This is fairly ok to compute for arbitrary $k$ and $q$, but quickly gets tough is $p$ increases. That's why selecting $p\le q$ is an advantage if the board is rectangular but not square.