Probability: there are $n$ rooms, and $m$ meetings, $m \leq n$, what's the probability of all meetings scheduled to a different room Quite new in stats... definitely not my strong area. I came across this probability question, and I am not sure how to do this! 
The question goes: 

pretend that there's this meeting scheduling engine used by this
  company and is not synced in real time, so when people schedule their
  meetings online to book a room, there may be overlaps. let's say there
  are $N$ rooms, and $M$ meetings, where $M \leq N$, what is the
  probability that all meetings scheduled to a different room?

My thought was that, the first meeting doesnt matter, can be in any room; then the 2nd meeting has $\frac{1}{N-1}$ chance of being in a room. so for two rooms not colliding, the chance of them being in separate rooms is $\frac{1}{N-1}$. Right? I am not confident about this one neither... 
Any hint/advice/guidance helps!
update
to clarify:

1) each room can only host up to one meeting
  2) one meeting can only happen in one room

 A: The total number of ways to distribute meetings to rooms is $N^M$, because for example the first meeting can take place in $N$ rooms, the second also in $N$ rooms, and so on.
Let's calculate the number of favorable situations. The first meeting has $N$ choices. The second has $N-1$ choices and so on. So the probability of all meetings are in different rooms is $$\frac{N(N-1)\cdots(N-M+1)}{N^M}$$
A: If you are not talking about scheduling (for which time overlap would
an issue), but instead talking about just where the meeting takes place, then
there are $N^M$ possible assignments of meeting rooms. There are $\binom{N}{M}$
ways to choose $M$ different rooms for the meetings and $M!$ ways to assign the $M$ meetings to those rooms, so the probability is ${\binom{N}{M}M! \over N^M}$.
A: Same solution as the others with a slightly different perspective (somewhat in line with the reasoning you tried to provide).
Let $E_i$ be the event that room $i$ does not collide with rooms $1,2,3,...,i-1$.
As you noted the first room does not matter.
$$P(E_1)=1$$
For the second room, there are $N-1$ rooms we can pick out of the $N$ total taking into account the first room was chosen:
$$P(E_2|E_1)=\dfrac{N-1}{N}$$
Similarly,
$$P(E_3|E_2,E_1)=\dfrac{N-2}{N}$$
And the pattern continues in this fashion. 
$$P(E_i|E_{i-1},E_{i-2},\cdots ,E_1) = \dfrac{N-i+1}{N}$$
You seek 
$$P(E_1\cap E_2\cap \cdots\cap E_m) = P(E_1)P(E_2|E_1)P(E_3|E_2,E_1)\cdots P(E_m|E_{m-1},E_{m-2},\cdots,E_1)$$
$$=1\cdot \dfrac{N-1}{N}\cdot \dfrac{N-2}{N}\cdots \dfrac{N-m+1}{N}$$
