How to not use stars and bars to solve unordered sampling with replacement?

Question

Say I want to find the number of different ways of unordered sampling k elements from n distinguishable elements with replacement, and a common way is to use the method of stars and bars. Although this is artful and easy to understand, I want to know whether there are other methods to solve this problem, and if there are, what are those methods? And the more direct, the rawer, the better!

BTW, I have searched in a few textbooks on this problem, and it seems all of them are using the above method.

In this answer Marko Riedel gives a proof using the Pólya enumeration theorem. — Brian M. Scott, Aug 16 '16 at 21:41

score 4 · Accepted Answer · answered Aug 16 '16 at 13:32

If you multiply out $(1+x+x^2+\ldots)^n$ distributively then you get a term $x^{k_1}x^{k_2}\>\cdots\> x^{k_n}$ for each independent sampling of a term in each of the factors. Collecting terms according to increasing powers of $x$ we obtain a formal series $\sum_{k\geq0} a_k\>x^k$, whereby $a_k$ counts the number of samplings $(k_1,k_2,\ldots,k_n)$ with $k_1+k_2+\ldots+k_n=k$.

Now $$(1+x+x^2+\ldots)^n={1\over (1-x)^n}=\sum_{k=0}^\infty{-n\choose k} (-x)^k$$ (binomial series). It follows that $$a_k=(-1)^k{-n\choose k}={n+k-1\choose k}\ .$$

Thanks for your nice solution using generating function! – Daniel Aug 16 '16 at 13:56 — Daniel, Aug 16 '16 at 13:56

How to not use stars and bars to solve unordered sampling with replacement?

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