Let the coordinates of points be:
From A(1, 1) to C(4, 6), the total steps right = \(4 - 1 = 3\), and up = \(6 - 1 = 5\), for a total of \(3 + 5 = 8\) steps.
So, the number of ways to reach C from A is: \[ \binom{8}{5} \]
From C(4, 6) to B(8, 10), the total steps right = \(8 - 4 = 4\), and up = \(10 - 6 = 4\), for a total of \(4 + 4 = 8\) steps.
So, the number of ways to reach B from C is: \[ \binom{8}{4} \]
Multiply the two: \[ \binom{8}{5} \times \binom{8}{4} = 56 \times 70 = \boxed{3920} \]
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: