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?

Looks like seats are identical.
So, first place a person any seat. this will act as a reference point. This can be done only in one way. Then seat remaining $(m-1)$ persons in remaining $(m+r-1)$ seats and this can be done in $\dbinom{m+r-1}{m-1}$ ways. Considering arrangements, multiply with $(m-1)!$.
So, 

$\dbinom{m+r-1}{m-1} \times (m-1)!$ ways

This is same as the answer given in the book because

$\dbinom{m+r-1}{m-1}(m-1)!=\dbinom{m+r-1}{r}(m-1)!$

