On a chessboard, a king is to be allowed to move one square at a time: horizontally to the right, vertically downward, or diagonally to the right and downward. Imagine a reduced $4\times 4$ chessboard, with the king beginning in the top-left square.
By how many routes can he reach the bottom-right square?
By how many routes can a similar journey be made on a full $8 \times 8$ chessboard?
Break into cases based on how many diagonal motions the king makes.
On a $4\times 4$ grid, if there are $k$ diagonal movements, there are $3-k$ rights and $3-k$ downs still needing to be made.
There are $\frac{(3-k)+(3-k)+k}{k!(3-k)!(3-k)!}$ ways to arrange $k$ $G$'s (representing diagonals) $3-k$ $R$'s (representing rights) and $3-k$ $N$'s (representing downs).
Summing over all cases gives an answer. Adjusting the numbers and number of cases allows for any size board or initial position.