With rectangular axes of coordinates, the number of paths from $(1,1)$ to $(8,10)$ via $(4,6)$, where each step from any point $(x, y)$ is either to $(x$, $y+1)$ or to ( $x+1, y$ ), is

When we move from point $(x, y)$ to either $(x, y+1)$ or $(x+1, y)$, we're essentially:

Moving right by 1 unit: $(x, y) \to (x+1, y)$

Moving up by 1 unit: $(x, y) \to (x, y+1)$

This is called a lattice path problem.

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Since we must pass through $(4,6)$, we can split our journey into two parts:

Part 1: $(1,1) \to (4,6)$

Part 2: $(4,6) \to (8,10)$

The total number of paths = (Paths in Part 1) × (Paths in Part 2)

Why multiply? For each way to reach $(4,6)$, we can combine it with any way to go from $(4,6)$ to $(8,10)$. This is the multiplication principle in counting.

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To go from $(1,1)$ to $(4,6)$:

Right moves needed: $4 - 1 = 3$ steps

Up moves needed: $6 - 1 = 5$ steps

Total moves: $3 + 5 = 8$ steps

Out of these 8 steps, we need to choose which 3 will be "right" moves (the remaining 5 will automatically be "up" moves).

Number of ways = ${{}}^{8}C_{3} = \dfrac{8!}{3! \times 5!} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

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To go from $(4,6)$ to $(8,10)$:

Right moves needed: $8 - 4 = 4$ steps

Up moves needed: $10 - 6 = 4$ steps

Total moves: $4 + 4 = 8$ steps

Out of these 8 steps, we choose which 4 will be "right" moves.

Number of ways = ${{}}^{8}C_{4} = \dfrac{8!}{4! \times 4!} = \dfrac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$

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Total number of paths = $56 \times 70 = 3920$

Key Insight: In lattice path problems, if you need to go right $r$ times and up $u$ times, the number of ways is ${{}}^{r+u}C_{r}$ (or equivalently ${{}}^{r+u}C_{u}$).