Hi Manika,
Could you please specify your doubt or the step at which you are finding difficulty in understanding our solution?

Hi,
There are 4 blocks which can be covered in 2 lefts and 2 down moves, thus 4!/2! *2!.

If there are 'm' horizontal way and 'n' vertical ways than number of ways of reaching the destination are (m+n)!/(m!*n!) In this case m = 2 and n=2

This is exactly similar to normal grid problem. Which part didn't you get? There's only one way to go from top left corner of square P to bottom left corner to square P. Number osf ways to go from A to top left corner of P is (4!/2!*2!) and the number of ways to go from bottom left corner to B is (6!/4!*2!). Hence we multiply.

Hi,
Can you elaborate which part of the solution you have trouble understanding? It will help us to help better.

please explain from start

Hi,
We have added a diagram in solution for more clarity. Kindly have a look at it.

Hi Shubham,
The number of ways he can select going through $$2$$ vertical and $$4$$ horizontal roads is given by $$(4+2)C_2 $$ = $$6!/(4!*2!)$$

Hi, to travel from A to O, Neelam has tot travel two horizontal and two vertical steps. (Total 4 steps). Out of these 4 steps, any 2 can be horizontal and remaining vertical. Hence, the number of ways of selecting these two steps = $$^4C_2$$ = $$\frac{4!}{2!\times\ 2!}$$. Similarly, we can calculate for Q to B. Hope that answers your query.

Hi Krishna,
The value of $$^{n+m}C_m$$ is same as the value of $$^{n+m}C_n$$. So you can calculate either $$^9C_2\ \ or\ \ ^9C_7$$ as both the values are the same.
$$^9C_2\ \ =\ \dfrac{9!}{7!\times\ 2!}$$ and $$^9C_7\ \ =\ \dfrac{9!}{7!\times\ 2!}$$
Hope this helps!

Got it sir , thank you