Probability of m repetitions when drawing from a uniform distribution When independently drawing $n$ random numbers from the set of positive integers $1, ..., M$ using a uniform distribution, I would like to know the probability of drawing any number $m$ or more times. More specifically, I would like to state a reasonable upper bound for this like 

"I am 99% sure no number will be drawn more than $m$ times"

Obviously, the probability of drawing a single number is $\frac{1}{M}$ and that of drawing a specific number $m$ times would be $\frac{1}{M^m}$, I think. However, I am trying to find the probability of drawing any number $m$ times.
 A: Firstly: the probability of drawing a specific number $m$ times is: $$\dbinom n m \dfrac{(M-1)^{n-m}}{M^n}$$
Secondly: the probability of drawing two specific numbers $m$ times each is $$\dbinom n {m,m,n-2m}\dfrac{(M-2)^{n-2m}}{M^n}$$
And so on.
Thirdly: Use the Principle of Inclusion and Exclusion.  The probability that some number is drawn exactly $m$ times is:
$$\sum_{k=1}^{\min\{M,\lfloor n/m\rfloor\}}(-1)^{k-1}\dfrac {M!}{k!(M-k)!}\cdot\dfrac{\;}{\qquad}\cdot\dfrac{}{\qquad}$$
Can you complete?
A: Too much for a comment:
Let $E_{i}$ denote the event that number $i\in\left\{ 1,\dots,M\right\} $
is drawn at least $m$ times.
Applying inclusion/exclusion we find:
$$P\left(\bigcup_{i=1}^{M}E_{i}\right)=\sum_{i=1}^{M}P\left(E_{i}\right)-\sum_{1\leq i<j\leq M}P\left(E_{i}\cap E_{j}\right)+\cdots$$
Applying symmetry we find:
$$P\left(\bigcup_{i=1}^{M}E_{i}\right)=\sum_{k=1}^{M}\binom{M}{k}\left(-1\right)^{k-1}P\left(E_{1}\cap\cdots\cap E_{k}\right)$$
I have not much hope to finding suitable expressions for $P\left(E_{1}\cap\cdots\cap E_{k}\right)$,
though we do have $P\left(E_{1}\cap\cdots\cap E_{k}\right)=0$ if
$km>n$. 
