>To be precise, there will be **1000!** bijections between them. But there will be one "best" bijection.

Just for clarity, John got that monstruous number by computing 1000! (that's 1000 factorial). Why is the number of bijections 1000! ? For any finite set, the number of bijections is the same as the number of permutations. In fact, permutations are bijections of a set with itself (AKA *automorphisms* in the category of sets). To see that the number of permutations is \\(n!\\) for a set with cardinality \\(n\\), consider a set \\(X\\). Pick an element \\(x\\), there's permutation that sends \\(x\\) to each of the \\(n\\) elements of \\(X\\). Once you've determined where \\(x\\) is mapped, pick another element \\(y\\). There's a permutation that sends \\(y\\) to each of the \\(n-1\\) remaining elements (you can't send it to the same element as \\(x\\) since permutations are bijections).

By induction, the set of permutations on \\(X\\) is \\(n(n-1)\ldots 1= n!\\).

Just for clarity, John got that monstruous number by computing 1000! (that's 1000 factorial). Why is the number of bijections 1000! ? For any finite set, the number of bijections is the same as the number of permutations. In fact, permutations are bijections of a set with itself (AKA *automorphisms* in the category of sets). To see that the number of permutations is \\(n!\\) for a set with cardinality \\(n\\), consider a set \\(X\\). Pick an element \\(x\\), there's permutation that sends \\(x\\) to each of the \\(n\\) elements of \\(X\\). Once you've determined where \\(x\\) is mapped, pick another element \\(y\\). There's a permutation that sends \\(y\\) to each of the \\(n-1\\) remaining elements (you can't send it to the same element as \\(x\\) since permutations are bijections).

By induction, the set of permutations on \\(X\\) is \\(n(n-1)\ldots 1= n!\\).