$\sum_{a+b+c=n}ab \binom{n}{a,b,c}=n(n-1)3^{n-2}$ 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.

$$
\begin{array}{l}
 \sum\limits_{a + b + c = n} {ab\left( \begin{array}{c}
 n \\  a,b,c \\ 
 \end{array} \right)}  = \sum\limits_{a + b + c = n} {ab\frac{{n!}}{{a!b!c!}}}  = \sum\limits_{\begin{array}{*{20}c}
   {1 \le a,b}  \\   {a + b + c = n}  \\
\end{array}} {\frac{{n!}}{{\left( {a - 1} \right)!\left( {b - 1} \right)!c!}}}  =  \\ 
  = \sum\limits_{\begin{array}{*{20}c}
   {0 \le c,d,e}  \\   {d + e + c = n - 2}  \\
\end{array}} {\frac{{n!}}{{d!e!c!}}}  = \frac{{n!}}{{\left( {n - 2} \right)!}}\sum\limits_{\begin{array}{*{20}c}
   {0 \le c,d,e}  \\   {d + e + c = n - 2}  \\
\end{array}} {\frac{{\left( {n - 2} \right)!}}{{d!e!c!}}}  =  \cdots  \\ 
 \end{array}$$
same scheme with the other.
