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

Attempt:
First, we notice that for the first box, we have $n$ choices, and for the second box we also have $n$ choices, and so on. Thus, we have $n^N$ ways to place the balls into the boxes. Pick $m$ balls out of the total $n$ balls, that gives ${n \choose m}$. In how many ways can these $m$ balls go into the first box? Well, in just ${1 \choose 1 }= 1 $ ways. Thus,
$$ P = \frac{ {n \choose m } }{n^N } $$
Is this correct?
 A: The number of ways to choose the $m$ items in the first box is $\binom{n}{m}$ and the number of ways to put the remaining $n-m$ items into the remaining $N-1$ boxes is $(N-1)^{n-m}$; therefore, the number of ways to put $m$ items into the first box is
$$
\binom{n}{m}(N-1)^{n-m}
$$
So the probability is
$$
\frac{\binom{n}{m}(N-1)^{n-m}}{N^n}=\frac{\binom{n}{m}}{N^m}\left(1-\frac1N\right)^{n-m}
$$
Note that the Binomial Theorem guarantees that
$$
\begin{align}
\sum_{m=0}^n\binom{n}{m}(N-1)^{n-m}
&=((N-1)+1)^n\\
&=N^n
\end{align}
$$
A: If we're just concerned with the first box, and all other boxes are equally likely, it's just a binomial with $n$ trials, for which $p=1/N$.
$$\Pr(m=k) \,=\, {n \choose k}\,p^{k}(1-p)^{n-k} \,=\, {n \choose k}\,N^{-k}\left(1-\frac1N\right)^{n-k}.$$
A: If $X$ is the number of balls that fall into the first box, then $P(X=m)$ follows the binomial distribution, where $p=1/N$.
A: I think that the problem is different if the balls are indistinguishable. If we have 2 balls and 3 boxes we have 6 ways.
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In this case we have $ \frac{(2+(3-1)!)}{2! (3-1)!}=6 $
