I'm looking at a sum of a function which involves a trinomial expansion with $$ \sum_{i+j+k = N, 0\leq\{i,j,k\}\leq N} \binom{N}{i,j,k} f(i,j,k) $$ I started by rewritting this with $$ \sum_{i,j,k = 0}^N \delta(N-i - j - k) \binom{N}{i,j,k} f(i,j,k) $$ and then in my problem I specifically need to get rid of one of these variables, so I tried to apply the delta function with $$ \sum_{i,j = 0}^N \Theta(N- i - j) \binom{N}{i,j,N-i-j} f(i,j,N-i-j) $$ where $\Theta$ is the Heaveside function and I'm defining $\Theta(0) = 1$.
Is this even correct? And if so, can these sums be written in a way that gets rid of the $\Theta$ function?
You are basically ensuring $i+j\leq N$ with the $\Theta$ function. Equivalently, you want to ensure that for a fixed $i$, you have $0\leq j\leq N-i$. With this, the sums look like
$$\sum_{i=0}^N\sum_{j=0}^{N-i}\binom{N}{i, j, N-i-j}f(i, j, N-i-j),$$
and you do not need the $\Theta$ function.