probability of $k$ boxes contain exactly $1$ ball Occupancy problem with balls and boxes. Suppose there are $N$ balls and $M$ boxes. The balls are thrown to the boxes at random. What is the probability of $k$ boxes contain exactly $1$ ball? where $k=1,2,...\min(N,M)$
 A: What is the chance there are exactly $k$ boxes with a single ball?
For $1\leq i\leq N$, define $A_i$ to be the event that the $i$th box gets exactly one ball. 
These are exchangeable events whose intersections have  multinomial probability 
$$P(A_1\cdots A_j)={M \choose 1,1,1,\dots,1,M-j}\left({1\over N}\right)^{1}\cdots
\left({1\over N}\right)^{1}\left(1-{j\over N}\right)^{M-j}$$ and so
by inclusion-exclusion, the probability that there are exactly $k$ boxes
with a single ball is
$$p(k)=\sum_{j=k}^{\min(M,N)} (-1)^{j-k} {j\choose k}{N\choose j}{M\choose j}
 \,j! \,\left({1\over N}\right)^j\left(1-{j\over N}\right)^{M-j}.$$
For example, according to this formula with $M=6$ balls and $N=8$ boxes, the chance that exactly six boxes get a single ball is 
$$p(6)={315\over 4096}={20160\over 8^6}\approx .07690,$$
exactly as in Marko Riedel's answer. 
A: The interpretation here  is that the balls are  distinguishable as are
the boxes. This gives the combinatorial species
$$\mathfrak{S}_{=M}(\mathcal{E} + \mathcal{U}\mathcal{Z}
+ \mathfrak{P}_{\ge 2}(\mathcal{Z})).$$
This  translates  into  the  bivariate generating  function  with  $z$
representing the  number of balls and  $u$ the number of  boxes with a
single occupant:
$$G(z, u) = (\exp(z)-z+uz)^M.$$
Observe that $$N! [z^N] G(z, 1) = N! [z^N] \exp(Mz) = M^N,$$
as it ought to be.
It  follows that the  count of  configurations with  $k$ boxes  with a
single occupant is
$$N! [z^N] {M\choose k} z^k (\exp(z)-z)^{M-k}
= N!  {M\choose k} [z^{N-k}] (\exp(z)-z)^{M-k}$$
for a probability of
$$M^{-N} N!  {M\choose k} [z^{N-k}] (\exp(z)-z)^{M-k}.$$
The inner term may be re-written with Stirling numbers to get
$$N!  {M\choose k}
[z^{N-k}] \sum_{p=0}^{M-k} {M-k\choose p} (\exp(z)-1)^p
(1-z)^{M-k-p}
\\ = N!  {M\choose k}
\sum_{p=0}^{M-k} {M-k\choose p} \times p! \\ \times
\sum_{q=0}^{N-k} \frac{1}{(N-k-q)!} {N-k-q\brace p} 
(-1)^{q} {M-k-p\choose q}.$$
The  preceding sum  may  be  subjected to  additional  algebra but  no
significant simplification was found by this writer.
The following  Maple code was used  to compare the  results from total
enumeration to the formulae presented above.

with(combinat, stirling2);

Q :=
proc(N, M)
    option remember;
    local ind, d, gf, mset, k, p;

    gf := 0;

    for ind from M^N to 2*M^N-1 do
        d := convert(ind, base, M);
        d := [seq(d[q], q=1..N)];

        mset := convert(d, `multiset`);

        k := 0;
        for p in mset do
            if p[2] = 1 then
                k := k + 1;
            fi;
        od;

        gf := gf + u^k;
    od;

    gf;
end;

EX :=
proc(N, M)
    local k, gf;

    gf := 0;

    for k from 0 to min(M,N) do
        gf := gf + u^k *
        N!*binomial(M,k)*coeftayl((exp(z)-z)^(M-k), z=0, N-k)
    od;

    gf;
end;

EX2 :=
proc(N, M)
    local k, gf;

    gf := 0;

    for k from 0 to min(M,N) do
        gf := gf + u^k *
        N!*binomial(M,k)*
        add(binomial(M-k,p)*p!*
            add(1/(N-k-q)!*stirling2(N-k-q,p)*
                (-1)^q*binomial(M-k-p, q), q=0..N-k), p=0..M-k);
    od;

    gf;
end;

For example the distribution of the number of single occupants for six
balls in eight boxes is
$$20160\,{u}^{6}+100800\,{u}^{4}+33600\,{u}^{3}+80640\,{u}^{2}+20496
\,u+6448.$$
The first coefficient is $20160$  because we must first choose the two
boxes  that   remain  empty  and  distribute  a   permutation  of  six
distinguishable balls into the remaining six boxes, for a total of
$${8\choose 2} \times 6! = 20160.$$
A: All the option to throw the balls is given by $CC^N _M$
The options in which k boxes contains exactly 1 ball is given by inclusion exclusion formula for the properties:
$p_i = $the i'th box contains exactly one ball.
then just divide those to get the probability
