Calculating the probability of the 'common birthday problem' differently yields a different result? Everyone is calculating the birthday problem here  by multiplying the probabilities of each person's birthday.  Like the following "first oe has $365/365$ the second has $364/365$..... and so on..." . And this is the complementary of the probability so you take $1$ minus it.
This is fine from a probability point of view. But once you go to combinatorics I get confused.
The thing that bothers me is that the formula you get in the end is similar to the combinatorial formula of choosing $k$ from $n$ if order matters divided by all possible selections of $k$.
My problem is that the formula should be calculated the same but when order doesnt matter.  That is, I want to choose $k$ days from a year of $365$ so I got ${365}\choose{k}$ all divided by $k+n-1  \choose  k$ which is the formula for choosing $k$ from $n$ with repetitions allowed and order doesn't matter.
I don't understand why it does not yield the same result. Even though this seems to me to be the correct way to solve it. Because I dont care about order, all what matters is that I choose k different days out of the year which is the complementary of the probability.

The number of ways we can pick $k$ days from $365$ is, of course, $365 \choose k$.
The number of ways that $N$ people can have their birthdays distributed over $k$ days, allowing for repetition, ignoring order, and requiring at least one on each day is $N-1 \choose k-1$.  This can be proven via induction on $k$.
The number of ways of ordering $N$ people, is, of course $N!$.
The number of ways $N$ people can have birthdays, counting different orders is $365^N$.
Thus, the probability that $N$ people have, between them, $k$ birthdays is:
$$p(k,N) = {{{365 \choose k} {N-1 \choose k-1} N! \over 365^N}} $$
For the case of $k=N$, this reduces to the known result.
