Combinatorics identity

Your calculation is correct.

Here is an algebraic proof based upon the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g. \begin{align*} [x^k](1+x)^n=\binom{n}{k} \end{align*}

We obtain for $a,b,c\geq 0$ and $0\leq n\leq a+b+c$

\begin{align*} \binom{a+b+c}{n}&=[x^n](1+x)^{a+b+c}\\ &=[x^n](1+x)^a(1+x)^b(1+x)^c\\ &=[x^n]\sum_{i=0}^a\binom{a}{i}x^i(1+x)^b(1+x)^c\tag{1}\\ &=\sum_{i=0}^n\binom{a}{i}[x^{n-i}]\sum_{j=0}^b\binom{b}{j}x^j(1+x)^c\tag{2}\\ &=\sum_{i=0}^n\sum_{j=0}^{n-i}\binom{a}{i}\binom{b}{j}[x^{n-i-j}]\sum_{k=0}^c\binom{c}{k}x^k\tag{3}\\ &=\sum_{i=0}^n\sum_{j=0}^{n-i}\binom{a}{i}\binom{b}{j}\binom{c}{n-i-j}\tag{4}\\ &=\sum_{{i+j+k=n}\atop{i,j,k\geq 0}}\binom{a}{i}\binom{b}{j}\binom{c}{k}\qquad\qquad\qquad\qquad\qquad n\geq 0\tag{5} \end{align*} and the claim follows.

Comment:

• In (1) we apply the binomial theorem.

• In (2) we use the linearity of the coefficient of operator and apply the rule \begin{align*} [x^{p-q}]A(x)=[x^p]x^qA(x) \end{align*} We also set the upper limit of the sum to $n$, since the exponent $n-i$ of $x$ has to be non-negative.

• In (3) we do same step with $(1+x)^b$ as we did in (2).

• In (4) we select the coefficient of $[x^{n-i-j}]$.

• In (5) we set $k:= n-i-j$ and sum over $i,j,k\geq 0$ by noting that $\binom{p}{q}=0$ if $q>p$.