Goggles in the locker room problem 
At a swimming pool, 5 swimmers leave their goggles in the locker room. The instructor finds the googles and randomly returns them to the swimmers. What is the probability that exactly one swimmer gets the correct goggles?

So I let the swimmers be $A, B, C, D, E$ and their corresponding goggles be $a, b, c, d, e$. I fixed $A$ to be the only one to get his/her goggle, $a$, and decided to multiply by $5$ later since it could happen to anyone. Then I just laid out all the cases for the remaining four people and goggles not to match, obtaining a total of $10$ cases. There are $5!$ possible cases in total, so I got $\frac{5}{12}$, but that is wrong... Any help would be appreciated. 
 A: Suppose we have already selected which of the five swimmers receives his or her own goggles.  That leaves four swimmers, none of whom will receive his/her own goggles.
There are $4!$ ways to distribute goggles to the remaining four swimmers.  From these, we must remove those arrangements in which one or more swimmers receives his/her own goggles.  
There are $\binom{4}{1}$ ways to select a swimmer to receive his/her own goggles and $3!$ ways to distribute the remaining goggles.  Hence, there are $\binom{4}{1}3!$ ways to distribute the goggles so that one of the remaining four swimmers receives his/her own goggles.
However, if we subtract $\binom{4}{1}3!$ from the total, we will have subtracted those arrangements in which two of the four swimmers each receive their own goggles twice, once for each way of designating one of those two swimmers as the swimmer who receives his or her own goggles.  We only want to subtract these cases once, so we must add them back.
There are $\binom{4}{2}$ ways to select two of the four swimmers as the recipients of their own goggles and $2!$ ways to distribute the remaining two goggles to the other swimmers.   Hence, there are $\binom{4}{2}2!$ arrangements in which two of the four swimmers receive their own goggles.
If we subtract those arrangements in which a swimmer receives his/her own goggles and then those arrangements in which two swimmers receive their own goggles from the total, we will not have excluded those cases in which three swimmers receive their own goggles.  This is because we subtracted those cases three times, once for each of the three ways we could designate one of those swimmers as the swimmer who receives his/her own goggles, then added them three times, once for each of the $\binom{3}{2}$ ways we could designate two of those three swimmers as the ones who receive their own goggles.  Consequently, we must subtract those cases in which three of the swimmers receive their own goggles.
There are $\binom{4}{3}$ ways to select three swimmers who receive their own goggles and $1!$ ways to distribute the remaining goggles to the remaining swimmer.  Hence, there are $\binom{4}{3}1!$ arrangements in which three of the swimmers receive their own goggles.
By subtracting those cases in which a swimmer receives his/her own goggles, adding cases in which two swimmers receive their own goggles, then adding cases in which three swimmers receive their own goggles, we have subtracted those cases in which all four swimmers receive their own goggles twice.  This is because we subtracted them four times, once for each way we could designate one of the four swimmers as the one who receives his/her own goggles; added them six times, once for each of the $\binom{4}{2} = 6$ times we could designate two of the four swimmers as the swimmers who receive their own goggles; and subtracted them four times, once for each of the $\binom{4}{3} = 4$ ways we could designate three of the four swimmers as the ones who receive their own goggles.  Since we only want to subtract cases in which all four swimmers receive their own goggles once, we must add them back.
There are $\binom{4}{4}$ ways for all four swimmers to receive their own goggles and $0!$ ways to distribute the remaining goggles to the remaining swimmers.  
Hence, by the Inclusion-Exclusion Principle, the number of ways that four goggles can be distributed to four swimmers so that none of them receive his/her own goggles is 
$$4! - \binom{4}{1}3! + \binom{4}{2}2! - \binom{4}{3}1! + \binom{4}{4}0!$$
Since there are five ways to choose the swimmer who receives his/her own goggles, the number of distributions of goggles in which exactly one of the five swimmers receives his/her own goggles is 
$$\binom{5}{1}\left[4! - \binom{4}{1}3! + \binom{4}{2}2! - \binom{4}{3}1! + \binom{4}{4}0!\right]$$
Since there are $5!$ ways to distribute the five goggles to the five swimmers so that each swimmer receives one pair of goggles, the desired probability is 
$$\frac{\binom{5}{1}\left[4! - \binom{4}{1}3! + \binom{4}{2}2! - \binom{4}{3}1! + \binom{4}{4}0!\right]}{5!}$$
A permutation in which no object is left in its original position is called a derangement.  By the Inclusion-Exclusion Principle, the number of derangements of $n$ objects is 
$$!n = D_n = \sum_{k = 0}^{n} (-1)^k\binom{n}{k}(n - k)!$$
In counting distributions in which none of the four swimmers receives his/her own goggles, we counted derangements of four objects.
A: Let the swimmers be $S = \{1,...,n\}$ and let the goggles be $G = \{1,...,n\}$. Goggle $i$ belongs to swimmer $i$. Every possible way the instructor gives back the goggles to the swimmers corresponds to a bijective function $$ \sigma :G \rightarrow S. $$
Since $G = S$, every $\sigma$ is a permutation of the set $S = \{1,...,n\}$. All permutations are equally likely to occur. The permutations that correspond to the event that exactly one swimmer gets his own glasses are the permutations with exactly one fixed point. A fixed point of a permutation $\sigma$ is an $i \in S$ such that $\sigma(i) = i$.  So the probability you are looking for is $$ \frac{\text{Number of permutations of n elements with exactly one fixed point}}{\text{Number of all permutations of n elements}}.$$ 
In total, there are $n!$ permutations of $n$ elements. It is a well-known result, that the number of permutations with exactly one fixed point is given by $$ n! \sum_{k=0}^{n-1} \frac{(-1)^k}{k!}. $$ Therefore, the probability that exactly one of the $n$ swimmers is getting his own goggles is given by $$ p_n = \sum_{k=0}^{n-1} \frac{(-1)^k}{k!}.  $$ In your case $(n=5)$, this results in $$ p_5 = 3/8. $$ In the limit, we obtain $$ \lim_{n \rightarrow \infty} p_n = 1/e. $$
