Say, $r = \text{Move Right}$ and $u = \text{Move Up}$
so using ${10}$ combination of $r$ and ${10}$ combinations of $u$ moves we get a solution.
Convert the graphical moves to text and one such solution we get,
$=\{u, u, u, u, u, u, u, u, u, u, r, r, r, r, r, r, r, r, r, r\}$
now all possible arrangements of them is given by = $\frac{20!}{10! \times 10!} = {20 \choose 10}$
now we need to discard the segment move from $(4,4)$ to $(5,4)$:
to do that we first calculate how many solutions to our problem to reach $(10, 10)$ involves that segment.
We'll then subtract those solutions from the total number of solutions.
Number of solutions to reach from$\left(0,0\right)$ to $\left(4,4\right),$
$=$all possible arrangements of $\{r, r, r, r, u, u, u, u\}$
$=\frac{(4+4)!}{4! \times 4!} = {8 \choose 4}$
definitely we take the segment $\left(4,4\right) \text{to} \left(5,4\right) = 1.$
now, Number of solutions to reach from $\left(5,4\right) \text{to} \left(10,10\right),$
$=$ all possible arrangements of $\{r, r, r, r, r, u, u, u, u, u, u\}$
$=\frac{(6+5)!}{6! \times 5!} = {11 \choose 5}$
so required number of solutions for Q.85 is given by option D
i.e. $={20 \choose 10} - {8 \choose 4} \times 1 \times {11 \choose 5}$