I was doing a question where I was distributing $15$ identical candy to four people, with the restriction that a specific person had at most 4 candy.
I did this in two ways: first way by subtracting the ways of having at least 5 candy for that person from the number of total unrestricted ways.
The second way was by counting the number of ways I can distribute the candy after giving that special person only 0 candy, 1 candy, ..., 4 candies.
Then I saw that it could be generalised to form some sort of identity by counting:
Number of non-negative integer solutions to
$$x_1+x_2+\ldots+x_m = n$$
where $m\leq n$, $x_i \geq 0$ for $i=1,2,\ldots,m-1$ and $x_m\geq a$ for $a<n$.
The first method I counted
$$\binom{n+m-1}{m-1} - \binom{n+m-a-2}{m-1}$$
and the second method I counted
$$\sum_{k=0}^a \binom{n+m-2-k}{m-2}.$$
Then $$\sum_{k=0}^a \binom{n+m-2-k}{m-2}=\binom{n+m-1}{m-1} - \binom{n+m-a-2}{m-1} .$$ Could this possibly be generalised even more and is this already a known identity?
