As shown in the figure, in an $8 \times 8$ square grid, there are two bugs at points $A$ and $B$ (the bug at point $A$ is called bug $a$, and the bug at point $B$ is called bug $b$). Each bug moves 1 grid (choosing one of the four directions: up, down, left, or right). If bug $b$ moves two grids, and bug $a$ moves three grids, the number of ways in which the distance from bug $b$ to point $A$ is less than or equal to the distance from bug $a$ to point $A$ is $\qquad$ ways.
【Analysis and Solution】
Counting, Multiplication Principle, Addition Principle, Permutations, Combinations.
(1) $b$ bug moves 2 steps:
$\rightarrow \rightarrow$, finally $b$ bug is 3 steps away from point $A$;
$\leftarrow \leftarrow$, finally $b$ bug is 1 step away from point $A$;
$\uparrow \uparrow$ or $\downarrow \downarrow$, finally $b$ bug is $\sqrt{5}$ steps away from point $A$ (diagonal of a $2 \times 1$ rectangle);
$\rightarrow \uparrow$ or $\rightarrow \downarrow$ and their permutations, finally $b$ bug is $\sqrt{5}$ steps away from point $A$ (diagonal of a $2 \times 1$ rectangle);
$\rightarrow \leftarrow$ and its permutations, finally $b$ bug is 1 step away from point $A$;
$\leftarrow \uparrow$ or $\leftarrow \downarrow$ and their permutations, finally $b$ bug is 1 step away from point $A$;
$\uparrow \downarrow$ and its permutations, finally $b$ bug is 1 step away from point $A$;
Finally, $b$ bug is 1 step away from point $A$, there are $1 + 4 \times P_{2}^{2} = 9$ ways;
Finally, $b$ bug is $\sqrt{5}$ steps away from point $A$ (diagonal of a $2 \times 1$ rectangle), there are $2 + 2 \times P_{2}^{2} = 6$ ways; Finally, $b$ bug is 3 steps away from point $A$, there is 1 way.
(2) $a$ bug moves 3 steps:
$\rightarrow \rightarrow \rightarrow$ or $\uparrow \uparrow$ or $\leftarrow \leftarrow \leftarrow$ or $\downarrow \downarrow \downarrow$, finally $a$ bug is 3 steps away from point $A$;
$\rightarrow \rightarrow \leftarrow$ or $\leftarrow \leftarrow \rightarrow$ or $\uparrow \uparrow \downarrow$ or $\downarrow \downarrow \uparrow$ and their permutations, finally $a$ bug is 1 step away from point $A$;
$\rightarrow \rightarrow \uparrow$ or $\rightarrow \rightarrow \downarrow$ or $\leftarrow \leftarrow \uparrow$ or $\leftarrow \leftarrow \downarrow$ or $\uparrow \uparrow \rightarrow$ or $\uparrow \uparrow \leftarrow$ or $\downarrow \downarrow \rightarrow$ or $\downarrow \downarrow \leftarrow$ and their permutations,
finally $a$ bug is $\sqrt{5}$ steps away from point $A$ (diagonal of a $2 \times 1$ rectangle);
$\rightarrow \uparrow \leftarrow$ or $\rightarrow \uparrow \downarrow$ or $\rightarrow \leftarrow \downarrow$ or $\uparrow \leftarrow \downarrow$ and their permutations, finally $a$ bug is 1 step away from point $A$;
Finally, $a$ bug is 1 step away from point $A$, there are $4 \times C_{3}^{1} \times C_{2}^{2} + 4 \times P_{3}^{3} = 36$ ways;
Finally, $a$ bug is $\sqrt{5}$ steps away from point $A$ (diagonal of a $2 \times 1$ rectangle), there are $8 \times C_{3}^{1} \times C_{2}^{2} = 24$ ways;
Finally, $a$ bug is 3 steps away from point $A$, there are 4 ways.
(3) Finally, the distance between $b$ bug and point $A$ is less than or equal to the distance between $a$ bug and point $A$:
(1) When finally $b$ bug is 1 step away from point $A$,

then finally $a$ bug is 1 step or $\sqrt{5}$ steps (diagonal of a $2 \times 1$ rectangle) or 3 steps away from point $A$,
there are $9 \times (36 + 24 + 4) = 9 \times 64 = 576$ ways;
(2) When finally $b$ bug is $\sqrt{5}$ steps away from point $A$ (diagonal of a $2 \times 1$ rectangle),

then finally $a$ bug is $\sqrt{5}$ steps (diagonal of a $2 \times 1$ rectangle) or 3 steps away from point $A$,
there are $6 \times (24 + 4) = 168$ ways;
(3) When finally $b$ bug is 3 steps away from point $A$,

then finally $a$ bug is 3 steps away from point $A$,
there are $1 \times 4 = 4$ ways;
In summary, the number of ways for the distance between $b$ bug and point $A$ to be less than or equal to the distance between $a$ bug and point $A$ is $576 + 168 + 4 = 748$ ways.