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I'd like a clarification about or some insight into one possible form of solution to the following problem:

  1. Suppose that each of n sticks is broken, into one long and one short part. The 2n parts are arranged into n pairs from which new sticks are formed. Find the probability (a) that the parts will be joined in the original order ...

Quite easily I came up with the correct solution $$\frac{2^n n!}{(2n)!}$$ by the reasoning that

  • all sticks can be ordered in $(2n)!$ ways and
  • $n$ original pairs of sticks can be ordered in $n!$ ways and each such pair can be further ordered in 2 ways.

According to answers this solution can also be written as $$\frac{2^n n!}{(2n)!} = [1 \cdot 3 \cdot 5 \cdot \dotso \cdot (2n-1)]^{-1} = (2n-1)!!^{-1}$$

which actually appears as the first candidate (before equal sign). I can't figure out how to interpret this form and by what reasoning could I arrive at it. Note I'm not trying to find out how is the middle derived from the left but rather how is the middle derived from the problem description.

The problem is from the book An Introduction to Probability Theory and Its Applications Vol.1 by W. Feller.

sqxmn
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    Surely in the equation ,you have the LHS upside down, it must be $\frac{(2n)!}{2^n\cdot n!}$. The probability can't be equal to the double factorial ! – true blue anil Jul 25 '15 at 10:13
  • Thank you for the spot. Fixed. In the book the double factorial is to the power of -1. – sqxmn Jul 25 '15 at 18:09

2 Answers2

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As noted in the comments,

$$\frac{2^nn!}{(2n)!}=(2n-1)!!$$

is clearly not the answer to the question, since in general it’s bigger than $1$ and so cannot be a probability. Rather, it’s the number of ways to form $n$ pairs from a set of $2n$ distinguishable objects. Only one of those pairings is correct, so the desired probability is

$$\frac1{\frac{2^nn!}{(2n)!}}=\frac{(2n)!}{2^nn!}\;.$$

You already found one explanation of how one might arrive directly at the result $(2n-1)!!$; you can also find it in this answer and both answers to this question.

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Brian M. Scott
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Same question with enlightening comment appeared when I was writing appropriate title: Combinatorial interpretation of double factorial. Leaving my question for the problem reference.

When all sticks are sorted and the process of pairing starts from the beginning there's only one way to pair them correctly and total number of ways to do such pairing is described in the aforementioned answer.

sqxmn
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