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 birthday probability can be written as:
$$\prod_{k=1}^{n-1}(1-k/365)=\frac{364!/(365-n)!}{365^{n-1}}=\frac{\binom{365}{n}}{365^{n}/n!}.$$
To use binomial coefficients, you can reason as follows. If you have $n$ people, you need to choose $n$ distinct birthdays, which can be done in $\binom{365}{n}$ ways. The total number of ways of picking birthdays, when order in which you pick the people doesn't matter, is $365^{n}/n!$. 
