Number of ways to go from $d_1$ to $d_8$ using only 7 King moves

Question

On an $8$ by 8 chess board, how many possible ways are there to move from
square $d_1$ (1st row from bottom and 4th column from left ) to
square $d_8$ (8th row from bottom and 4th column from left)?

You can only use 7 king moves i.e. you can move one square per move, and there are 3 ways to do so: horizontally, vertically and diagonally.
For example - $d_1, d_2, d_3, d_4, d_5, d_6, d_7, d_8$

Answer 1

I accidentally interpreted letter as row and number as column. Just rotate and/or replace the corresponding words to match your interpretation of how letters/numbers correspond to directions)

You should be able to convince yourself that the king must always go to the right, to the up-right, or to the down-right on every turn in order to arrive at d8 by the seventh move. You can use more exciting techniques which generalize easily, but with how small the numbers here are, you can map out all possibilities using a grid and putting as numbers in the grid the number of distinct paths which led to that location.

Such a grid might begin as such:

$\begin{bmatrix} & & & 1 & 4\\ & & 1 & 3 & 10 & \cdots\\ & 1 & 2 & 6 & 16\\ 1 & 1 & 3 & 7 & 19\\ & 1 & 2 & 6 & 16 & \cdots\\ & & 1 & 3 & 10\\ & & & 1 & 4 \\ & & & & 1 & \cdots \end{bmatrix}$

Create the next number in the next column by adding the number directly to the left, to the left-down and to the left-up. for example the third row fifth column entry $16$ was formed by doing $3+6+7$ and shows there are $16$ different paths you can take from $d_1$ to $c_5$ using exactly four king-moves.

Keep in mind the physical boundaries of your board so that you don't move off the edge, i.e. keep your grid as an $8\times 8$. Any time you try to add a number outside of the grid or not filled in, treat it as a zero.

Answer 2

Basically you can only make three types of moves: going up, up-left or up-right. Also, the steps of going up-left must be the same as that of going up-right.

Conditioned on the steps $k$ of going up, which can only be 1, 3, 5 and 7, we have the number of ways would be $\binom{7}{k} \binom{7-k}{(7-k)/2}$. I calculated the result to be 393.