Hi Aditya Nayak,
It is given in the question that, "Each step from (x,y) can be only upon either (x+1,y) OR (x,y+1)"
which means that you can move one unit either ONLY along X-axis or ONLY Y-axis from that particular point (x,y)
Eg, If you start from (1,1), then you can either go to (2,1) OR (1,2) and no other point from here.
If you chose (2,1) from there you can go only to (3,1) or only (2,2).
Also, note that you can only go forward- So if we consider the case given by you
If he starts from (1,1) he can reach (9,1) but how will reach (4,6) without coming backwards and there is no condition given for backward movement
Only forward movement is possible, so from (1,1) we have to reach (4,6) and then to (8,10)
I hope this helps!

Hi Apurv,
In the given problem, we have calculated the number of paths from A(1,1)  to C (4,6) is $$^8C_3$$ = 56
And, number of paths from C(4,6) to B (8,10) is $$^8C_4$$ = 70
From A to C - We need to take 3 steps in the x direction and 5 steps in the y direction
From C to B - We need to take 4 steps in the x direction and 4 steps in the y direction
But you are saying it is $$^7C_3$$ and $$^7C_4$$. Could you please tell me how you got this, and please check and elaborate on your doubt so I can clear your doubt
Regards

Each step from (x,y) can be only upon either (x+1,y) OR (x,y+1) means that you can move one unit either ONLY along X-axis or ONLY Y-axis from that particular point (x,y)

Eg: If you start from (1,1) then you can either go to (2,1) OR (1,2) and no other point from here.
If you chose (2,1) the from there you can go only to (3,1) or only (2,2).
Hope this clarifies.

Hi Pranov,
You can take any $$^8C_3$$ or $$^8C_5$$, as the value of both are same
$$^8C_3=\ \dfrac{8!}{\left(8-3\right)!3!}=$$ 56
$$^8C_5=\ \dfrac{8!}{\left(8-5\right)!5!}=$$ 56
You can refer to this property
$$^nC_r=^nC_{n-r}$$
I hope this helps!