Number of paths of length 16 exist between (0,0) and (7,9)

Question

What Is the number of paths of length 16 exist between (0,0) and (7,9) on the 2D-axis, such there is no path that intersect with $y=x-3$?

• Every path is one step right [from (a,b) to (a+1,b)] or one step up [from (a,b) to (a,b+1)]

Thanks.

Answer 1

This might not be the shortest path, but that's how I would go for solving this exercise:

Any paths of length 16 from $(0,0)$ to $(7,9)$ can be seen as an arrangement of $7$ red balls and $9$ blue balls: red ball = "go right", blue ball = "go up".

So there are $16!$ combinations of these balls, but as we can not distinguish to blue balls, we have to divide this number by $7!$. Same for red balls. We obtain in total: $$ N=\frac{16!}{7!\cdot9!}$$ possible paths of length 16 between $(0,0)$ and $(7,9)$.

Now, we have to count the number $M$ of paths that cross the line $y=x-3$ so that the number we are looking for is given by $N-M$.

Note that any path of lengths $16$ which crosses this line have to pass by one of the following points: $x^1=(3,0)$, $x^2=(4,1)$, $x^3=(5,2)$, $x^4=(6,3)$ or $x^5=(7,4)$. Each of these paths can be written as the union of two paths, one path $\gamma_1$ going from $(0,0)$ to $x_i$ and another path $\gamma_2$ going from $x_i$ to $(7,9)$. In particular, since the length of $\gamma_1\cup \gamma_2$ is $16$, we know that the length of $\gamma_1$ must be of length $x^i_1+x^i_2$ (where $x^i = (x_1^i,x_2^i)$) and the length of $\gamma_2$ must be of length $16-(x^i_1+x^i_2)$. So you can apply the same idea as above and count the number of path crossing $x^i$ for $i=1,...,5$.