i tried to identify the coefficients $\gamma n \nu \mu $ in
$$(a+b+c)^n = \sum_{\nu=0}^{n} \sum_{\mu = \nu}^{n} \gamma n \nu \mu ~a^\nu b^{\mu- \nu}c^{n- \mu}.$$
I used the Binomial Theorem, but didn't succeed, can you help me?
Binomial Theorem: $$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k$$
$$ \begin{align} (a+b+c)^n &= \sum_k\binom nka^k(b+c)^{n-k} \\ &= \sum_k\binom nka^k\sum_l\binom klb^lc^{n-k-l} \\ &= \sum_k\sum_l\binom nk\binom kla^kb^lc^{n-k-l} \\ &= \sum_{k+l+m=n}\binom nk\binom kla^kb^lc^m \\ &= \sum_{k+l+m=n}\binom n{k,l,m}a^kb^lc^m \qquad\text{where $\binom n{k,l,m}\overset{\rm def}=\binom nk\binom kl=\frac{n!}{k!l!m!}. $} \end{align} $$ Such products of binomial coefficients are called multinomial coefficients. The coefficent $\binom n{k,l,m}$ counts permutations of the letters of the word aaa...bb...ccc... of length $n$ with $k$ letters 'a', with $l$ letters 'b', and with $m$ letters 'c'; for instance $\binom7{2,1,4}$ counts permutations of aabcccc. Such permutations (with repetitions) are the products you get when working out $(a+b+c)^n$, leaving monomials as they are obtained from applying the distributive law, without regrouping the same variables together.