# Is my argument wrong? (A combinatorial exercise)

How many ways are there to arrange $m$ distinct flags on a row of $r$ flagpoles? The order of the flags on the flagpoles (from top to bottom) matters.

My argument is: I have $mr$ points and I have to decide where to put the $m$ flags, so the result should be $\binom{mr}{m}$. But the second point of the exercise let me think that the right answer might be $m(m+1)\cdots(m+r-1)$ or $r(r+1)\cdots(r+m-1)$.

Is my argument wrong? And if it is, where is the mistake?

-

What I would do is introduce the flags to the layout in a fixed order (say, alphabetical by the official French name of the country). For each flag, I have the option of either placing at the top of one of the $r$ flagpoles, or placing it immediately below one of the flags I have already positioned. The total number of different choices I can make is then $$r(r+1)(r+2)\cdots(r+m-1) = \frac{(r+m-1)!}{(r-1)!}$$
It seems you are right. My reasoning was: I have $mr$ position for the $m$ flags, some of this position will be empty, I have to choose $m$ of those positions that are not-empty. And the result is $\binom{mr}{m}$, where is the mistake? –  Alex M Sep 5 '11 at 21:58
Why do you have $mr$ positions? –  gary Sep 5 '11 at 22:08