Closed form for a sum Please, i need help with this example, step by step.
Calculate the value of the next summation, i.e. express simple formula without the sum:
$$\sum_{n_1 + n_2 + n_3 + n_4 = 5} \frac{6^{n_2-n_4} (-7)^{n_1}}{n_1! n_2! n_3! n_4!}$$

I think the formula is

$\sum_{n1 + n2 + n3 + n4 = n} = \frac{n!* x^{n_1} * x2^{n_2} * x3^{n_3}* x4^{n_4}}{n2! n2! n3! n4!}$
I don't know how to proceed, i stuck here.
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Hint.  The multinomial theorem is
$$(x_1 + x_2 + \ldots + x_k)^n = \sum_{n_1 + \cdots + n_k = n} {n \choose n_1, \ldots, n_k} x_1^{n_1} x_2^{n_2} \cdots x_k^{n_k}$$
Now let $k=4$, $n=5$ and find values of $x_1,x_2,x_3,x_4$ to make the right-hand side summand look a bit like the summand in your expression.  Note that $${n \choose n_1, \ldots, n_k} = \frac{n!}{n_1! \cdots n_k!}$$ so you get the $n_1!n_2!n_3!n_4!$ in the denominator for "free"...
