expanding $(a+b+c+...)^x$ where $x$ is natural neatly Suppose that $a,b,c,d,...$ are unknown variables.
One wishes to expand $(a+b+c..)^x$ where $x$ is natural number ina neat manner (for e.g. using combination, sigma etc.).
What would be some way?
Also, what would be the number of terms that would allow writing neatly?
 A: One can expand $(x_1+x_2+x_3+\dots +x_m)^n$ by using the multinomial theorem. It states that 
$$(x_1+x_2+x_3+\dots +x_m)^n=\sum_{k_1+k_2+\dots+k_m=n}\binom{n}{k_1,k_2,k_3,\dots,k_m}\prod_{t=1}^mx_t^{k_t}$$
$$=\sum_{k_1+k_2+\dots+k_m=n}\binom{n}{k_1,k_2,k_3,\dots,k_m}x_1^{k_1}x_2^{k_2}x_3^{k_3}\cdots x_m^{k_m}$$
Where $$\binom{n}{k_1,k_2,k_3,\dots,k_m}=\frac{n!}{k_1!k_2!\cdots k_m!}$$
What this basically means is take all the sets $\{k_1,k_2,\dots,k_m\}$ such that it's elements add up to $n$ and each element is a whole number, then find $\binom{n}{k_1,k_2,k_3,\dots,k_m}$ for each of those sets, multiply the resultant by $x_1^{k_1}x_2^{k_2}x_3^{k_3}\cdots x_m^{k_m}$ and add it all together.
For example: $(a+b+c+d)^3$. The coefficient of $a^2b$ will be $$\binom{3}{2,1,0,0}=\frac{3!}{2!1!0!0!}=\frac{6}{2\cdot1\cdot1\cdot1}=3$$
While the coefficient of $abc$ will be
$$\binom{3}{1,1,1,0}=\frac{3!}{1!1!1!0!}=\frac{6}{1\cdot1\cdot1\cdot1}=6$$
Doing this for the rest of the terms we get 
$$a^3+3 a^2 b+3 a b^2+b^3+3 a^2 c+6 a b c+3 b^2 c+3 a c^2+3 b c^2+c^3+3 a^2 d+6 a b d+3 b^2 d+6 a c d+6 b c d+3 c^2 d+3 a d^2+3 b d^2+3 c d^2+d^3$$
Note that in the expansion you can see smaller binomial expansions (e.g. the first 4 terms $a^3+3 a^2 b+3 a b^2+b^3$ is just $(a+b)^3$). In fact $(x_1+x_2+x_3+\dots +x_m)^n$ will contain every binomial expansion $(x_i+x_j)^n$ where $i\ne j$ and $1\le i,j\le m$. This follows from the fact that you can choose a set which contains exactly 2 non-zero elements $k_i,k_j$ such that $k_i+k_j=n$.
