Consider an $m \times n$ grid of unit squares, indexed by $(i, j)$ with $1 \leq i \leq m$ and $1 \leq j \leq n$.
Then why is the number of possible paths from $(1,n)$ to $(m,1)$ such that you are only going down and to the right equal to $\binom{m+n-2}{m -1}$?
could someone give me some intuition for this?
This question is asked in reference to B1 question of the 2023 Putnam competition, as I do not understand how the solutions derived this result.
The number of rights you would have to take along the path would be $m-1,$ and the amount of downs you would have to take along the path would be $n-1.$ This means that the number of paths we can possibly take is the same as the number of ways to arrange $DDD...DDDRRR...RRR$ where we have $m-1$ Rs and $n-1$ Ds. This is equal to $C((n-1)+(m-1),m-1)=C(n+m-2,m-1)$ because all rights and all downs are indistinguishable.