Probability Distribution of n candidates There are n candidates in a conference, the organizers have created name tags for each of them but accidentally gave them out at random.
I am taking distribution as X = how many candidates got the correct name tags (correct means their own name tag)
X: 0 1 2 3 4 5 . . . . . n
P: ? ? ? ? ? ? ? ? ? ? ? ?
n = no. of candidates who got their own name tag which will of course be equal to no. of candidates
Is it possible to find it for n candidates or will have to go individually?
(By individually I mean by taking n = 2 (it will mean there are only 2 candidates), it's pretty easy to find with n = 2,3 but after that things get complicated.)
 A: We want to know the probability that a random permutation of $n$ objects has exactly $k$ fixed points, where $0 \le k \le n$.  In terms of the original problem statement, this is the probability that exactly $k$ of the candidates get their correct name tag.
A permutation with no fixed points is called a derangement.  It is well-known that the number of derangements of $n$ objects is
$$D_n = n! \sum_{i=0}^n (-1)^n \frac{1}{i!}$$
See the Wikipedia article on derangements for a proof of this formula using the inclusion-exclusion principle.
There are $n!$ permutations of $n$ objects, all of which we assume are equally likely.  We would like to count the number with exactly $k$ fixed points.  There are $\binom{n}{k}$ ways to pick the $k$ objects, and then the permutation of the remaining $n-k$ objects must be a derangement.  So the number of permutations with exactly $k$ fixed points is
$$\binom{n}{k} D_{n-k}$$
and the probablity of such a permutation is
$$\frac{1}{n!} \binom{n}{k} D_{n-k} = \frac{1}{k!} \sum_{i=0}^{n-k} (-1)^i \frac{1}{i!}$$
after a little algebra.
A: Since the probability of any individual member getting their correct name tag is $\frac{1}{n}$, the total expected value considering all $n$ members is $n\frac{1}{n}$ = $1$.
