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?
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 $k_i+k_j=n$.
