<s>[INST] Probability of an array having all distinct numbers Suppose you have an array of size $2n$. There are two times $2n^2$ distinct numbers that can be put into the array without replacement, i.e. for each choice of number, there are two copies, so a maximum of two cells in the array can have the same value. What is the probability that all $2n$ numbers in the array are distinct?
This is what I came up with: The probability that the second number is distinct from the first is $(4n^2 - 2)/(4n^2 - 1)$, the probability that the third number is distinct from the first and the second, is $(4n^2 - 4)/(4n^2-2)$, and so on...Conditioning gives
$$
\frac{4n^2 - 2}{4n^2 - 1}\frac{4n^2 - 4}{4n^2-2}... \frac{4n^2 - 2(2n-1)}{4n^2 - 2n+1}
$$
I'm not sure if this is correct though.
 [/INST] Your solution is correct, though it may be incomplete. I'll start with what you have, including the $\frac{4n^2}{4n^2}$ term:
$$\frac{(4n^2)(4n^2-2)(4n^2-4)\cdots(4n^2-2(2n-1))}{(4n^2)(4n^2-1)(4n^2-2)\cdots(4n^2-(2n-1))}$$
Removing a factor of $2$ from each term in the numerator gives:
$$2^{2n}\times\frac{(2n^2)(2n^2-1)(2n^2-2)\cdots(2n^2-(2n-1))}{(4n^2)(4n^2-1)(4n^2-2)\cdots(4n^2-(2n-1))}\tag{1}$$
Notice now that the numerator is $\frac{(2n^2)!}{(2n^2-2n)!}$ and the denominator is $\frac{(4n^2)!}{(4n^2-2n)!}$
Let's multiply each by $\frac{(2n)!}{(2n)!}$ and see what happens:
$$\frac{(2n)!}{(2n)!}\frac{(2n^2)!}{(2n^2-2n)!}=(2n)!\frac{(2n^2)!}{(2n)!(2n^2-2n)!}=(2n)!\binom{2n^2}{2n}$$and:$$
\frac{(2n)!}{(2n)!}\frac{(4n^2)!}{(4n^2-2n)!}=(2n)!\frac{(4n^2)!}{(2n)!(4n^2-2n)!}=(2n)!\binom{4n^2}{2n}$$
Substituting these into $(1)$ gives:
$$2^{2n}\times\frac{(2n)!\binom{2n^2}{2n}}{(2n)!\binom{4n^2}{2n}}=2^{2n}\times\frac{\binom{2n^2}{2n}}{\binom{4n^2}{2n}}$$
</s>