Arrange $m$ people in $m+r$ seats around a round table. The answer is, according to the book: $(m-1)!\cdot \binom{m+r-1}{r}$
I get why this is true. You arrange $m$ people in their seats in $(m-1)!$ ways and then you put $r$ empty spots inbetween them.I was thinking - choose $m$ seats out of $m+r$ to put the people on => $\binom{m+r}{m}$
Arrange them once you've chosen the seats $\to (m-1)!$
So in total: $\binom{m+r}{m} \cdot (m-1)!$
This is obviously not true. What's wrong with the way I'm thinking?

With your way, you over count. If one group of $m$ seats that you choose, on the table, is a rotated version of another group of $m$, then you can get the same order, by just rotating them.
