Lattice Paths that Avoid a Point

Solution 1:

You need to look at the complement, which is all the paths from $(0,0)$ to $(7,5)$ that pass through $(3,3)$.

As you noted, the number of paths from $(0,0)$ to $(3,3)$ is ${6\choose3} = 20$, and the number of paths from $(3,3)$ to $(7,5)$ is ${6\choose4} = 15$. So the number of paths from $(0,0)$ to $(7,5)$ that pass through $(3,3)$ is all combinations of the above paths, which means $20\cdot15=300$ paths.

$792-300=492$, and so $492$ is the final answer.

Solution 2:

Your idea to subtract the number of paths passing through $(3,3)$ from $792$ was a good idea, but you didn't carry it out quite right. The number of paths that pass through $(3,3)$ is not $\binom{6}{3}$ nor $\binom{6}{4}$, but $\binom{6}{3}\binom{6}{4}$. This is because every path from $(0,0)$ to $(7,5)$ through $(3,3)$ consists of two "mini-paths," one from $(0,0)$ to $(3,3)$ and one from $(3,3)$ to $(7,5)$. There are $\binom{6}{3}$ ways to choose the first path and $\binom{6}{4}$ ways to choose the second, resulting in $\binom{6}{3}\binom{6}{4}$ ways to choose the entire path, and $792-\binom{6}{3}\binom{6}{4}$ paths avoiding $(3,3)$.