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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?

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    What do you have written down so far? – 2017-02-04
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    $$\begin{array}{|c|c|c|c}\hline1 &1 & 1 & \ldots\\ \hline 1 & \color{blue}{3} & \color{blue}{5} & \ldots\\\hline1 & \color{blue}{5} & \color{red}{8} & \ldots\\\hline\ldots & \ldots & \ldots & \ldots\end{array}$$ $$\color{blue}{3} = 1 + 1 + 1,~~\color{blue}{5} = \color{blue}{3} + 1 + 1,~~\color{red}{8} = \color{blue}{5} + \color{blue}{5} + \color{blue}{3}$$ – 2017-02-04
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    don't we need to count movements like 1->3->8->5-> ... – 2017-02-04
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    @Kiran We can't go up, to the left or diagonally up left. If we could then there would be infinitely many paths to any cell. – 2017-02-04
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    For some reason $8$ is equal to $5+5+3$. Wow, I'm bad at arithmetic today. – 2017-02-04
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    @Winther, thanks, got the point. – 2017-02-04
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    At least you noticed it before it was pointed out. The question of which method (*yours or mine*) is easier to manage is up to the user. They both require a little bit of theory to understand why they work, yours requires only addition but requires a great deal more information tracking/storing (*possibility for minor arithmetic mistakes as evidenced*), whereas mine requires no information storing but requires more complicated arithmetic involving multinomials and multiplication. – 2017-02-04

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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.

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    does it count the cases like, first to the 'right-bottom diagonal square', second to the 'right-bottom diagonal square', third to the 'top square' etc? (using the diagram given by Oliver821 in his comment, the movement i referred is 1 -> 3 -> 8 -> 5 -> ...) I was getting it too complex while thinking about the to and forth movements, though without repeating same square again – 2017-02-04
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    @Kiran I do not understand your comment. Not only are movements not described like you write but positions on the board are not described like you write so I'm not sure what you might be referring to. My answer leads the user to describe any path via a sequence of $G$'s, $R$'s and $N$'s and counts how many sequences exist that correspond to a path. For example the sequence $GGG$ corresponds to just traveling along the diagonal from start to end whereas $RRRNNN$ corresponds to traveling along the top edge followed by the right edge, etc... See https://en.wikipedia.org/wiki/Lattice_path – 2017-02-04
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    @Kiran you will never move from an 8 to a 5 because that would be moving left or up, neither of which are valid moves. The king only moves towards the goal. The only valid movements are down, right, or diag (down and right). No lefts, no ups, no diagonals using left or up are allowed. – 2017-02-04
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    thanks for clarifying. it is clear now. – 2017-02-04