Trinomial expansion with power constraints For the trinomial expansion $(a+b+c)^n$, I'd like to sum up the terms like $a^i b^j c^k$ with the constraint $i>j$. How to calculate it efficiently?
 A: Use first the binomial power expansion
\begin{align}
(\hat{a}+c)^n=\sum_{k=0}^n{{n}\choose k}\hat{a}^kc^{n-k}.
\end{align}
Now substitute $\hat{a}\equiv a+b$ to obtain
\begin{align}
(\hat{a}+c)^n&=\sum_{k=0}^n{{n}\choose k}\hat{a}^kc^{n-k}\\
&=\sum_{k=0}^n{{n}\choose k}(a+b)^kc^{n-k}
\end{align}
Now, apply the binomial expansion a second time to$(a+b)^k$, ie $(a+b)^k=\sum_{l=0}^k{{k}\choose l}a^lb^{k-l}$,
\begin{align}
(\hat{a}+c)^n&=\sum_{k=0}^n{{n}\choose k}\hat{a}^kc^{n-k}\\
&=\sum_{k=0}^n{{n}\choose k}(a+b)^kc^{n-k}\\
&=\sum_{k=0}^n{{n}\choose k}\biggl(\sum_{l=0}^k{{k}\choose l}a^lb^{k-l}\biggr)c^{n-k}\\
&=\sum_{k=0}^n\sum_{l=0}^k{{n}\choose k}{{k}\choose l}a^lb^{k-l}c^{n-k}
\end{align}
Your above condition to only sum up terms with $i>j$ translates in my notation to $l>k-l$ or equivalently $l>\tfrac{k}{2}$, thus your power expansion ultimately reduces to
\begin{align}
\sum_{k=0}^n\sum_{l>\frac{k}{2}}^k{{n}\choose k}{{k}\choose l}a^lb^{k-l}c^{n-k}
\end{align}
A: I'll do a small example:
$(a+b+c)^3=((a+b)+c)^3$
by associative property of addition. expanding this gives:
$$(a+b)^3+3(a+b)^2c+3(a+b)c^2+c^3$$
extracting from the binomial expansions $\lfloor\frac{k+1}{2}\rfloor$ terms that have $i>j$ in $(a+b)^k$, gives:
$$a^3+3a^2b+3a^2c+3ac^2$$ which is $$(a+c)^3-c^3+3a^2b$$
Multinomial theorem, simply gives a way to calculate the coefficients/full expansion quickly.
EDITS done to expand last paragraph, and fix a few stupid math errors.
A: Two variations.
We have 
\begin{align*}
(a+b+c)^n&=\sum_{k=0}^n\binom{n}{k}(a+b)^kc^{n-k}\\
&=\sum_{k=0}^n\binom{n}{k}\sum_{j=0}^k\binom{k}{j}a^jb^{k-j}c^{n-k}
\end{align*}
It follows
\begin{align*}
\color{blue}{\sum_{k=0}^n\ \sum_{j=\left\lfloor \frac{k}{2}\right\rfloor+1}^k \binom{n}{k} \binom{k}{j}a^jb^{k-j}c^{n-k}}
\end{align*}
We have
\begin{align*}
(a+b+c)^n&=\sum_{{i+j+k=n}\atop{i,j,k\geq 0}}\binom{n}{i,j,k}a^ib^jc^k
\end{align*}
It follows
\begin{align*}
\color{blue}{\sum_{{i+j+k=n}\atop{i>j,k\geq 0}}\binom{n}{i,j,k}a^ib^jc^k}
\end{align*}
