Prove a number is even using the Pigeonhole Principle Let n be an odd integer and let f be an [n]-permutation of length n, where [n] is the set of integers 1, 2, 3,...n
Show that the number 
x = (1-f(1))*(2-f(2))*...*(n-f(n))

is even using the pigeonhole principle
In this case, I don't understand what this function f is. What is an [n]-permutation of length n? Take f(2) for example. Permutations of [2] would be 1,2 and 2,1. So the way the problem is worded, f(2) must equal 12 or 21. If that's correct, which one? Will this number x still be even regardless of which [n]-permutation f(n) is?

Observe that
$$\sum_{i=1}^n (i - f(i)) = 0$$
Thus, the sum of an odd number of terms is $0$, so there exists at least one even term in this sum. Hence,
$$2 \mid \prod_{i=1}^n (i - f(i)).$$
