Probability of complete pairings from permutation of 14 people 14 people put their coats in a coatroom and are given back their coats completely at random. What is the probability that 7 pairs of people have received their coats switched?
(For example, if there were only 4 people, Harry, Ron, Hermione, and Tom: Harry’s and Tom’s coats are switched and Ron’s and Hermione’s coats are switched.)
Answer:
There are 14! ways of assigning the coats, all equally likely. There are
${14 \choose 2,2,2,2,2,2,2}$ ways of dividing the people into 7 pairs.
the people into 7 distinct pairs, and therefore
${14 \choose 2,2,2,2,2,2,2} \dfrac{1}{7!}$. By the law of equally likely outcomes, the required probability is: $\dfrac{{14 \choose 2,2,2,2,2,2,2} \dfrac{1}{7!}}{14!} = \dfrac{1}{2^7 7!}$
I was wondering, since the $\dfrac{1}{2^7 7!}$ looks so simple, was there another way to think about the question that also arrived at the $\dfrac{1}{2^7 7!}$ ?
 A: "Is there another way to think about the question that also arrived at the $\frac{1}{2^77!}$?"
Yes.
You should be able to convince yourself that it doesn't matter what order people arrive to claim their coats... all that matters is who receives what coat.  Let's assume the people are able to be alphabetically ordered.
Let the first person go and claim a coat.  They don't receive their correct coat with probability $\dfrac{13}{14}$.  Supposing this happens, let them look at the nametag on their coat and tell that person it is their turn to grab a coat.  That person grabs the first person's coat with probability $\dfrac{1}{13}$.
Now, given that this has all happened so far, the next remaining person earliest in the alphabet who hasn't picked up their coat yet goes to grab a coat.  They don't get theirs with probability $\dfrac{11}{12}$ and they again call the person whose coat they did grab to go in for their turn to grab a coat which happens to be this third person's coat with probability $\dfrac{1}{11}$.
Continuing in this fashion, given each outcome has occurred in a way such that we result in pairs of people whose coats were swapped, (i.e. that the permutation of coats can be written in disjoint cyclic form consisting only of $2$-cycles with no stationary points) we have as a probability:
$$\dfrac{13}{14}\times\dfrac{1}{13}\times\dfrac{11}{12}\times\dfrac{1}{11}\times\cdots \times \dfrac{3}{4}\times\dfrac{1}{3}\times\dfrac{1}{2}\times\dfrac{1}{1}$$
or simplified
$$\dfrac{1}{14\times 12\times 10\times\cdots\times 2}=\dfrac{1}{14!!}=\dfrac{1}{2^7\cdot 7!}$$
