$\sum_{a+b+c=n}ab \binom{n}{a,b,c}=n(n-1)3^{n-2}$ I want to prove $\sum_{a+b+c=n}ab \binom{n}{a,b,c}=n(n-1)3^{n-2}$.
Considering the multinomial theorem, setting two of the variables equal to $1$ and finally differentiating both sides twice, we get another identity, which is $\sum_{a+b+c=n}a(a-1) \binom{n}{a,b,c}=n(n-1)3^{n-2}$, which looks different from the former.
Actually I also proved the former identity combinatorially but I don’t know how these two different looking results are related to each other.
 A: Another approach reducing multinomial coefficients to binomial coefficients.
$$
\begin{align}
\sum_{a+b+c=n}ab\binom{n}{a,b,c}
&=\sum_{a=1}^n\sum_{b=1}^{n-a}ab\binom{n}{a}\binom{n-a}{b}\tag{1a}\\
&=\sum_{a=1}^n\sum_{b=1}^{n-a}a(n-a)\binom{n}{a}\binom{n-a-1}{b-1}\tag{1b}\\
&=\sum_{a=1}^na(n-a)\binom{n}{a}2^{n-a-1}\tag{1c}\\
&=\sum_{a=1}^na(n-1)\binom{n}{a}2^{n-a-1}-\sum_{a=1}^na(a-1)\binom{n}{a}2^{n-a-1}\tag{1d}\\
&=\sum_{a=1}^nn(n-1)\binom{n-1}{a-1}2^{n-a-1}-\sum_{a=1}^nn(n-1)\binom{n-2}{a-2}2^{n-a-1}\tag{1e}\\
&=\frac{n(n-1)}2\,3^{n-1}-\frac{n(n-1)}2\,3^{n-2}\tag{1f}\\[9pt]
&=n(n-1)\,3^{n-2}\tag{1g}
\end{align}
$$
Explanation:
$\text{(1a):}$ when $n=a+b+c$, $\binom{n}{a,b,c}=\binom{n}{a}\binom{n-a}{b}$
$\text{(1b):}$ $b\binom{n-a}{b}=\binom{n-a-1}{b-1}$
$\text{(1c):}$ sum in $b$ using the Binomial Theorem
$\text{(1d):}$ $n-a=(n-1)-(a-1)$
$\text{(1e):}$ $a\binom{n}{a}=n\binom{n-1}{a-1}$
$\phantom{\text{(1e):}}$ apply once to the left sum and twice to the right sum
$\text{(1f):}$ sum in $a$ using the Binomial Theorem
$\text{(1g):}$ simplify
A: $$
\begin{array}{l}
 \sum\limits_{a + b + c = n} {ab\left( \begin{array}{c}
 n \\  a,b,c \\ 
 \end{array} \right)}  = \sum\limits_{a + b + c = n} {ab\frac{{n!}}{{a!b!c!}}}  = \sum\limits_{\begin{array}{*{20}c}
   {1 \le a,b}  \\   {a + b + c = n}  \\
\end{array}} {\frac{{n!}}{{\left( {a - 1} \right)!\left( {b - 1} \right)!c!}}}  =  \\ 
  = \sum\limits_{\begin{array}{*{20}c}
   {0 \le c,d,e}  \\   {d + e + c = n - 2}  \\
\end{array}} {\frac{{n!}}{{d!e!c!}}}  = \frac{{n!}}{{\left( {n - 2} \right)!}}\sum\limits_{\begin{array}{*{20}c}
   {0 \le c,d,e}  \\   {d + e + c = n - 2}  \\
\end{array}} {\frac{{\left( {n - 2} \right)!}}{{d!e!c!}}}  =  \cdots  \\ 
 \end{array}$$
same scheme with the other.
