probability question, n guests in a bar put on shoes leaving the bar. n guests in a bar  put on shoes leaving the bar. Each guest cannot differentiate the difference between a left shoe and a right shoe, nor their own shoe ! Find the probability that each guest has put on their own pair of shoes. Answer (i don't get why there is need for the odd members in the product whatsoever.) $${2\over 2n}{1\over 2n-2}{2\over 2n-3}...={2^n\over(n!)^2}.$$ BTW this is what it says in the book i think there is a typo with the second member,i think it being $1\over 2n-1.$
 A: If, as you say, the guests know nothing about the shoes (i.e. they put them on while blindfolded or something), then there are basically $2 n$ "identical" shoes waiting to be put on.
Now, the probability that the first guest will put on the correct pair of shoes is $\displaystyle \frac 2 {2 n} \frac 1 {2 n - 1}$. Why is this so? The first term accounts for the probability that he will pick either the left or the right shoe of his own pair, while the second term is the probability that he will pick the remaining shoe of his own pair. You can easily continue in this manner, formally using the product rule, to obtain:
$$
\underbrace{\frac 2 {2 n} \frac 1 {2 n - 1}}_{\text{first guest}} \underbrace{\frac 2 {2 n - 2} \frac 1 {2 n - 3}}_{\text{second guest}} \cdots = \frac {2^n} {(2n)!}
$$
A: I have similar problem:
Each of $n$ guests chooses a pair of shoes from $n$ pairs of shoes. What is probability that each guest chooses one right and one left shoe (not necessarily of each pair)?
Answer: 
$p=\frac {2^n}{\binom{2n}{n}}$.
