$$(x_1+x_2+...+x_m)^n=\sum_{(k_1 + k_2 +... +k_m) \ = \ n} {n \choose k_1,k_2...k_m} \prod^m_{t=1}x_t^{k_t}$$ Let's do $(a+b+c)^3$. That means $a =x_1, b =x_2, c=x_3=x_m$. The multiindex below the sigma function is the different $P_m$ partitions of $n$, consisting of summands $\ge0$. If $n =3$, then those partitions will be $3+0+0$, $2+1+0$ and $1+1+1$. That means that the following operations after the sigma needs to use all of those k-values. So, the summation will look like this: $${3\choose 3,0,0} \times a^3 \times b^0 \times c^0$$ $$+$$ $${3 \choose 2,1,0} \times a^2 \times b^1 \times c^0$$ $$+$$ $${3 \choose 1,1,1} \times a^1 \times b^1 \times c^1$$ $$=a^3 + 3a^2b + 6abc$$ But that's obviously wrong, as it's missing $b^3 + c^3 + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b$. Thing is, I can't see how the formula above is to be used any other way that the erroneous way I used it. This is how I've interpreted it:
Since the x has the same index as it's exponent, the order is stuck, which means instead of going through every combination of the $x_i^{k_i}$, it will only get the first $x$ raised to the first $k$th power, the second $x$ raised to the second $k$th power, and so on. This misses a lot of the summands. How have I misinterpreted the notation to produce this wrong algorithm?
Order is important for these partitions. So you would need to have a term for every partition below:
Think of it as saying the more cumbersome
$$\sum_{\substack{(k_1,\cdots,k_m) \in \Bbb Z_{\ge 0}^m \\ k_1 + \cdots + k_m = n}}$$