A function is selected at random from all the functions of the set A = {1,2,...,n} in to itself. The probability that the function selected is one-to-one is?

Question

anonymous · Answer

How many functions are there? How many of them are one to one?

anonymous · Answer

A one to one function from a finite set A to itself is also a bijection. So it is a permutation. How many permutation you can have in the the set A={1,2,3,... n}?

anonymous · Answer

So there are n! permutations.

anonymous · Answer

Possible answers: 1/n^n 
2/(n-1)!
1/n!
(n-1)!/n^(n-1)

Okay, I got it. You have n! of cases for the one-to-one correspondence. And since, the possible set of events in the set in to itself, where each member of the domain maps to a member (though not necessarily all) of the co-domain: this event set is n^n. 

P(one-to-one given into) = n!/n^n = (n-1)!/n^(n-1)

anonymous · Answer

P(one-to-one given into) = n!/n^n = (n-1)!/n^(n-1)
That's the right answer.