Birthday problem: no unique birthday (all collide)

Question

I have another variation of a birthday problem to solve. Given $n$ people and $m$ birthdays (let's keep in generic...), what would be the probability that none of the people has a unique birthday date? I've already run out of ways to approach the problem...

Answer 1

The Inclusion-Exclusion principle gives a formula of sorts. Start with a basic $m^n$ possible sets of birthdays.
We want to subtract the number of ways Person1 has a unique birthday. He has one of $m$ birthdays, and everyone else must fit into $m-1$ days. That is $m(m-1)^{n-1}$ ways. In fact, subtract that $n$ times, because there are $n$ people who might have a unique birthday.
But if two people have unique birthdays. There are $m\choose2$ ways to choose the days and $n\choose2$ ways to choose the people. They match up in $2!$ ways. We subtracted that twice (once for A and once for B) and need to add it back in once.
If three people have unique birthdays, we have already subtracted three times (for A,B,C), added back in three times (for AB,AC,BC) and need to subtract off again.
In the end, it comes to $$\frac{m^n-mn(m-1)^{n-1}+2!{m\choose2}{n\choose2}(m-2)^{n-2}-3!{m\choose3}{n\choose3}(m-3)^{n-3}+...}{m^n}$$ EDIT: Here is an approximation for large $n$ and $m$.
The number of people with a given birthday is a Poisson variable with parameter $\lambda=n/m$, so the probability it is not $1$ is $1-\lambda e^{-\lambda}$. So the chance that no birthday is unique is $$\left(1-\frac nme^{-n/m}\right)^m$$ This goes from $0$ to $1$ when $ne^{-n/m}\approx1$, or $m\approx n/\ln n$