You have N people at a dinner party. How likely is it that two guests have the same birthday?
The first insight is that the probability $A$ of having a birthday pair is one, less the probability $B$ of having $N$ totally unique birthdays. That is, $A=1-B$. Let's work $B$, it makes for easy counting.
We're looking for a probability, which in this case is a ratio between two things. The first thing we want to know is, how many ways are there for $N$ people to have birthdays (any birthdays at all)? That's easy: $365^N$. The other thing we want to know is, how many ways are there for N people to NOT share a birthday? That is, how many ways can we have $N$ unique birthdays? Well, for more than 365 people we must have at least one pair, but for less than this a little reflection reveals the answer to be $365 * 364 * 363 * 362 * ... * (365-N)$. In other words, let one person have any birthday, the next can have almost any birthday, the next one less, etc... The ratio of these is the probability we're after.
$$ P(N) = \frac{365 * 364 * 363 * ... * (365-N)} {365^N} = \frac{365!}{365^N (365-N)!} $$
