Probability of a duplicate after K tries out of combination of N If I have a total of n different items to choose from (example, sides of a die), and I continually choose an item from the set (example, roll the die), is there a way to calculate the odds of having hit a duplicate after having choosen k elements (example, k die rolls) without using factorials?
I've worked it out to the following:
\begin{equation}n!/(n-k)! n^k\end{equation} 
The problem is that this calculation becomes unruly with big numbers. If I was interested in the odds of hitting a duplicate from a set of 10 million after 500 draws, then this would equate to:
\begin{equation}10000000!/(9999500!*10000000^{500})\end{equation}
Is this kind of equation realistic to calculate? A colleague quoted such a probability one time and I was curious how he had calculated it.

You can calculate $\frac{n!}{(n-k)!n^k}$ with the following pseudocode :
x:=1
For j=1,...,k
x:=x * (n+1-j)/n

Then, $x$ has the desired value. This should work even for very large numbers.
