combinatorics - 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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Answer

We consider the multinomial theorem in the form
$$\begin{align*} \left(x_1+x_2+x_3+x_4\right)^5&=\sum_{{n_1+n_2+n_3+n_4=5}\atop{n_j\geq 0, 1\leq j\leq 4}} \binom{5}{n_1,n_2,n_3,n_4}x_1^{n_1}x_2^{n_2}x_3^{n_3}x_4^{n_4}\\ &=\sum_{{n_1+n_2+n_3+n_4=5}\atop{n_j\geq 0, 1\leq j\leq 4}}\frac{5!}{n_1!n_2!n_3!n_4!}x_1^{n_1}x_2^{n_2}x_3^{n_3}x_4^{n_4}\tag{1} \end{align*}$$

We obtain applying (1)
$$\begin{align*} \color{blue}{\sum_{{n_1+n_2+n_3+n_4=5}\atop{n_j\geq 0, 1\leq j\leq 4}}}& \color{blue}{\frac{6^{n_2-n_4} (-7)^{n_1}}{n_1! n_2! n_3! n_4!}}\\ &=\frac{1}{5!}\sum_{{n_1+n_2+n_3+n_4=5}\atop{n_j\geq 0, 1\leq j\leq 4}}\frac{5!}{n_1! n_2! n_3! n_4!} (-7)^{n_1}6^{n_2}1^{n_3}\left(\frac{1}{6}\right)^{n_4} \\ &=\frac{1}{5!}\left(-7+6+1+\frac{1}{6}\right)^5\\ &\,\,\color{blue}{=\frac{1}{933\,120}} \end{align*}$$