In the expansion of $(x_1+x_2+x_3+...+x_k)^n$ , the coefficient of $(x_1^{n_1} x_2^{n_2} ... x_k^{n_k} )$? Does this problem mean we are selecting $n_1$ of $x_1$, $n_2$ of $x_2$ ...so on Therefore, $C(n,n_1,n_2..n_k)$= ?? I am unable to formulate the logic

If $n_1 + n_2 + \dots + n_k \neq n$, then the coefficient is zero.
Otherwise, we use the multinomial coefficient:
$$\frac{n!}{n_1!n_2!\dots n_k!}$$
for the coefficient of $x_1^{n_1} x_2^{n_2} \dots x_k^{n_k}$.
How did I come to this answer?
A good way to think about $(x_1 + x_2 + \dots + x_k)^n$ is to visualize it as
$$\underbrace{(x_1 + x_2 + \dots + x_k) \dots (x_1 + x_2 + \dots + x_k)}_{n \text{ times}}.$$
Thus, in the expansion, for each term we are picking one term from each of the $(x_1  + x_2 + \dots + x_k)$ and then multiplying it together.
So, to arrive at $x_1^{n_1} x_2^{n_2} \dots x_k^{n_k}$ we must have picked $x_1$ $n_1$ times, $x_2$ $n_2$ times, and so forth. The coefficient of the term is then "how many ways is it possible to pick the terms like this? Which is simply an application of combinatorics, namely the multinomial coefficient.
