Number of ways of distributing $N$ balls into $M$ bins such that at least one bin has at least $n$ balls in it? 
What is the number of ways of distributing $N$ indistinguishable balls into $M$ bins such that at least one bin has at least $n$ balls in it?


My attempt:
The number of ways of of placing $N$ balls into $M$ bins is $\binom{N+M-1}{N}$. 
I tried, by stars and bars, to calculate the number of ways of distributing $N$ indistinguishable balls into $M$ bins such that exactly one bin has exactly $n$ balls in it: 
If the first or last bin contains the $n$ balls, we have used one partition, so there are $M-2$ left. This gives $2\binom{N-n+M-2}{N-n}$ ways.
If the second to $(M-1)$th bin contains the $n$ balls, we have used two partitions, which gives $(M-2)\binom{N-n+M-3}{N-n}$ ways. 
So the total number of ways of distributing $N$ indistinguishable balls into $M$ bins such that exactly one bin has exactly $n$ balls in it is 
$$2\binom{N-n+M-2}{N-n}+(M-2)\binom{N-n+M-3}{N-n}.$$
Is this correct? It seems like a strange result. 
Then I thought about simply summing this expression for $n$ going from $n$ to $N$, i.e.,
$$\sum_{k=n}^N 2\binom{N-k+M-2}{N-k}+(M-2)\binom{N-k+M-3}{N-k},$$ 
in order to obtain the expression for "at least $n$ balls," but I feel like this would be over-counting somehow. 
And then there is the issue of "at least one bin," which I am rather blunted by. 
Any help is much appreciated!
Note that I'm looking for a closed form solution to the problem in the yellow box. Thanks.

Seems that it's not correct, since you going to have double counting for the case where cell 1 and cell 2 both have $n$ balls.
Try use inclusion-exclusion principle.
It's easy using the properties $p_i = \mathrm{The\space ith\space cell\space has\space at\space least\space n\space balls}$ to calculate $E(0)$ the case when no property holds.
For $r$ different properties to hold, we want that $r$ specific cells will have at least $n$ balls we first put $n$ balls in each of them and then the rest $N - rn$ we split however we want, so we get: $$ w(p_{i_1},...,p_{i_r}) = {{N + M - rn - 1}\choose N - rn}$$ and with the number of ways to pick $r$ different properties and the fact that the properties are symmetric we get $$ w(r) = {M \choose r} \cdot  {{N + M - rn - 1}\choose N - rn} $$ 
Now we have that $$E(0) = {{N+M-1}\choose{N}} - \sum_{r=1}^M (-1)^r\cdot w(r)$$ 
After calculating $E(0)$ use the fact that whats youre loking for is the complement of $E(0)$ which gives you that the desired number is: $${{N+M-1}\choose{N}} - E(0) = \sum_{r=1}^M (-1)^r\cdot w(r)$$
