Date: 2008-06-02
Update: 2021-01-05 (format for the web)
Suppose we have $n$ numeric variables ($v_1,\dots,v_n$) and want to raise their sum to a certain exponent $r$. We can multiply the exponent away, expanding our “power of sum of variables” to be a summation.
\[\left(\sum_{i=1}^{n}v_{i}\right)^{r}=(v_{1}+v_{2}+\cdots+v_{n})^{r} = \sum_{k=1}^{A}\left(c_{k} \prod_{i=1}^{n} v_{i}^{r_{k,i}}\right)\]Each term in the expansion has a different combination of multiples of the variables $v_1,\dots,v_n$ such that those multiples add up to $n$.
\[\forall k: \sum_{i=1}^{n} r_{k,i}=r\]Therefore $A$, the number of terms, is equal to the number of ways to make $r$ unordered copies of $n$ distinct elements. This choice without order that allows repeats can be written with the multiset coefficient as “$n$ multichoose $r$”. It can also be written as a binomial coefficient or directly with factorials.
\[A={n+r-1 \choose r}=\frac{(n+r-1)!}{r!(n-1)!}\]As for the coefficients, we know that their sum is equal to $n^r$ (obvious when you consider the case of $(1+\cdots+1)^r$).
\[\sum_{k=1}^{A} c_{k}=n^{r}\]The coefficient of any given term is $n!$ divided by the product of the factorials of the variables’ exponents in that term.
\[c_{k}=\frac{n!}{\prod_{i=1}^{n} r_{k,i}!}\]If there are just two variables, the coefficients make up the $r$th row of Pascal’s Triangle (if the top is the $0$th row). So, when there are two variables, the binomial coefficient can be used: $c_{k}={n \choose r_{k,1}}$.