I am solving the question:
How large must a class be to make the probability of finding two people with the same birthday at least 50%?
The first solution I came up with is rather simple. It's based on finding $N$ people such that any pair amongst the $N$ people have different birthdays. This can be simply solved as multiplying probabilities of $N$ people have different birthdays. The first person has a probability of 1 of having a different birthday. The 2nd person has a probability of (364/365) of having a different birthday from the first person. The 3rd has a probability of (363/365) of having a different birthday from the first 2 people, and so on.
$$ \frac{365}{365}\frac{364}{365}\cdots\frac{365-N+1}{365} < \frac{1}{2} \\ = \frac{^{365} P_N}{365^N} $$ It turns out $N=23$. This is the correct answer based on what I saw on Google.
I'm now trying to think of this problem in terms of combinatorics. So I first started with thinking of 365 distinguishable objects into $N$ bins without replacement. Order doesn't matter, so this is combinations, and we get $\binom{365}{N}$. Now I want to find the number of combinations of 365 birthdays into $N$ bins WITH replacement, and this is simply $\frac{(365+N-1)!}{N!(365-1)!}$. So then I was thinking the probability of less than half of getting $N$ people with different birthdays is then
$$ \frac{\binom{365}{N}}{\frac{(365+N-1)!}{N!(365-1)!}} < \frac{1}{2} $$
But if I plug in $N=23$, I don't get the $\approx \frac{1}{2}$ that is expected. I get $\approx \frac{1}{4}$. What is wrong with my thinking using the combinations approach?