placing balls into compartments 
Suppose we have $n$ balls that are randomly distributed into $N$
compartments. Find the probability that $m$ balls will fall into the
first two compartments. Assume all $N^n$  arrangements are equally
likely.

Try.
First, we select $m$ balls out of $n$, this we can do in ${n \choose m}$ ways and we want to put them in the first two compartments. That is, we have ${n \choose m}$ choices for both compartments. Therefore, we can do this in ${n \choose m}^2$ ways. Now, the remaining $n-m$ balls must be placed in the remaining $N-2$ compartments. This we can do in $(N-2)^{n-m}$ ways. So probability must be
$$ \frac{ {n \choose m}^2 \cdot (N-2)^{n-m} }{N^n } $$
However, I have some doubts.
If we assume that $m$ are to be distributed to any $2$ compartments, not just the first $2$, then we would have an extra ${N \choose 2}$ term in the above, is this correct?
 A: There are $\binom{n}{m}$ ways to select which $m$ of the $n$ balls will be in the first two compartment and $2^m$ ways to distribute them to those two compartments.  The remaining $n - m$ balls can be distributed to the remaining $N - 2$ compartments in $(N - 2)^{n - m}$ ways.  Hence, the number of favorable cases is 
$$\binom{n}{m}2^m(N - 2)^{n - m}$$
Since there are $N^n$ ways to distribute the balls, the probability that exactly $m$ of the $n$ balls land in the first two of the $N$ compartments is
$$\frac{\binom{n}{m}2^m(N - 2)^{n - m}}{N^n}$$
Sanity Check:  Suppose $N = 5$ and $n = 3$.  Then we should have $$\Pr(m = 0) + \Pr(m = 1) + \Pr(m = 2) + \Pr(m = 3) = 1$$
Our formula gives 
\begin{align*}
\Pr(m = 0) & = \frac{\binom{3}{0}2^0(5 - 2)^{3 - 0}}{5^3} = \frac{1 \cdot 1 \cdot 27}{125} = \frac{27}{125}\\
\Pr(m = 1) & = \frac{\binom{3}{1}2^1(5 - 2)^{3 - 1}}{5^3} = \frac{3 \cdot 2 \cdot 9}{125} = \frac{54}{125}\\
\Pr(m = 2) & = \frac{\binom{3}{2}2^2(5 - 2)^{3 - 2}}{5^3} = \frac{3 \cdot 4 \cdot 3}{125} = \frac{36}{125}\\
\Pr(m = 0) & = \frac{\binom{3}{3}2^3(5 - 2)^{3 - 3}}{5^3} = \frac{1 \cdot 8 \cdot 1}{125} = \frac{8}{125}
\end{align*}
so 
$$\Pr(m = 0) + \Pr(m = 1) + \Pr(m = 2) + \Pr(m = 3) = \frac{27 + 54 + 36 + 8}{125} = 1$$
as expected.
