Solve $$\sum_{n_1+n_2+n_3=n}n_1\binom{n}{n_1,n_2,n_3}$$
My approach: The general term for the given expression is $$n_1*\frac{n!}{n_1!n_2!n_3!}$$
which gives $$\frac{n!}{(n_1-1)!n_2!n_3!}$$
Now, if $n_1 + n_2 + n_3 = n$ then $(n_1-1) + n_2 + n_3 = n-1$
Therefore, if we denote $n_1 – 1 $ by $n_k$ the given expression can be changed to $$n\sum_{n_k+n_2+n_3=n-1}\binom{n-1}{n_k,n_2,n_3}$$
Using the result for sum of multinomial coefficients, this gives $n*3^{n-1}$
I may be dead wrong about this. Any help is appreciated.
Solve the summation $\sum_{n_1+n_2+n_3=n}n_1\binom{n}{n_1,n_2,n_3}$
combinatoricsmultinomial-coefficients
Best Answer
This is a combinatorial proof:
We want to divide $n$ people into three groups $A,B,C$. Then we want to elect a leader from group $A$. The number of ways to do it is given by your first expression
$$ \sum_{n_{1}+n_{2}+n_{3}=n}{n_{1}\cdot\binom{n}{n_{1},n_{2},n_{3}}} $$
There is another way to do it; we first choose a leader from $n$ people and this person also automatically join group $A$. Then the remaining $n-1$ people can join any of the three groups. The number of ways to do it is given by your second expression
$$ n\cdot 3^{n-1} $$
Since we count the same object (only the methods differ), both expression must be equal