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.
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I am just absolutely clueless about this problem as how to start it. Please help me out.
Best Answer
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.