# Combinatorics problem regarding two kinds of elements

I'm trying to understand this problem as exposed in William Feller's An introduction to probability theory and its applications:

(d) Flags of one or two colors. In example (1.f) it was shown that r flags can be displayed on n poles in N = n(n+ 1) ... (n+r-1) different ways. We now consider the same problem for flags of one color (considered indistinguishable). Numbering the flags of such a display yields exactly r! displays of r distinguishable flags and hence r flags of the same color can be displayed in N/r! ways.

I understand that a group of r distinguishable elements yield r! ordered samples of size r. But I do not understand why is it that r undistinguished flags can be displayed in N/r! ways. As I think about it, since they are supposed to be undistinguished, shouldn't there be only one way to order them? Does N/r! always yield 1? If so, why?

Many thanks

Put stickers with numbers $$1$$ to $$r$$ on the flags. Now after displaying the $$r$$ flags on $$n$$ poles, you can rearrange the stickers in $$r!$$ ways. That is, $$r!$$ different ways to display distinguishable flags correspond to $$1$$ way to display indistinguishable flags. Hence, if there are $$N$$ ways to display distinguishable flags, then there are $$N/r!$$ ways to display indistinguishable flags. The number $$N/r!$$ is not always $$1$$. For example if there are $$n=2$$ poles and $$r=3$$ indistinguishable flags there are $$4$$ ways to display them (indexed by the number of flags on the first pole). In fact displaying $$r$$ indistinguishable flags on $$n$$ poles counts the number of ordered number partitions of $$r$$ into $$n$$ non-negative parts, in the example with $$n=2$$, $$r=3$$ the displays correspond to the $$4$$ partitions $$0+3, 1+2, 2+1, 3+0.$$