[wging]

Not so. Given a set of N elements from which you choose k (with order not mattering, since you only care whether a given two are the same) the number of pairs is N!/(k!(N-k)!). So it's (6000000!)/(2!(6000000-2)!).

We can take advantage of the cancellation properties of the factorial to write this as N\(N-1)\...\(N-k+1)/k! . If k=2 this resolves to N(N-1)/2.

You could also derive it like this: for ordered* pairs, you have N choices for the first and N-1 for the second (since you don't pair anything with itself but all other choices are permitted). This counts each pair exactly twice: each pair and its reverse are the same if you don't care about order. So N(N-1)/2.

Still trillions of combinations. Not 5999999! though.