After $n$ sticks are broken into two parts each, they are joined again randomly. Find the probability of them being joined in a certain way Each of $n$ sticks are broken into a longer and a shorter part. Out of these $2n$ parts, $n$ sticks are formed again by joining any 2 parts randomly. Find the probability that
a) The parts will be joined in the original order.
b) Each longer part will be paired with a shorter part.
Progress
I am just absolutely clueless about this problem as how to start it. Please help me out.
 A: Pick up the sticks an join the first picked up with the second, the third picked up with the fourth, et cetera.
The probality that the second picked up is the original match of the first is $\frac{1}{2n-1}$. If this has occurred then we go on with $2n-2$ sticks.
This leads to a probability of $\frac{1}{2n-1}\times\frac{1}{2n-3}\times\cdots\times\frac{1}{3}\times\frac{1}{1}=\frac{2^{n}n!}{\left(2n\right)!}$
for the event that the parts will be joined in original order.
Likewise reasoning we find probability $\frac{n}{2n-1}\times\frac{n-1}{2n-3}\times\cdots\times\frac{2}{3}\times\frac{1}{1}=\frac{2^{n}n!n!}{\left(2n\right)!}=2^n\binom{2n}{n}^{-1}$ for the event that every longer part will be paired with a shorter part.
A: A slightly different approach to the above problem:
(i) Assume there are $n$ indistinguishable baskets where each basket would hold exactly two pieces out of the total $2n$ number of sticks. The no. of ways this could be accomplished if the baskets were distinguishable would be $\frac{2n!}{(2!)^{n}}$.
Now, in order to find the total no. of pairings of different pieces of sticks we need to consider the baskets as indistinguishable which would result in a total of $\frac{2n!}{(2!)^{n}n!}$ different pairings each occurring with the same probability. The desired pairing as asked in (i) is unique thus the probability of its occurrence would then be given by $\frac{1}{\frac{2n!}{(2!)^{n} n!}}$ or simply $\frac{(2!)^{n} n!}{(2n)!}$
(ii) This part is almost similar to part (i) except for the fact that the total no. of pairings possible for the desired event is $n!$ thus resulting in the probability $\frac{(2!)^{n} n! n!}{(2n)!}$ or upon simplifying further $\frac{(2!)^{n}}{2n \choose n}$.
