Joining Long and Short Stick Suppose each of ten sticks is broken into a long part and a short part. The twenty parts are arranged into ten pairs and glued back together so that again there are ten sticks. What is the probability that each long part will be paired with a short part?
My solution is the following:
From these 20 shorter sticks, 2 sticks can be glued together in ${}_{20} C_{2}$ ways.
Now, there are 10 short sticks and 10 long. Let's mark them as $\text{S1,S2,....,S10 }$ for short sticks and $\text{L1,L2,....,L10 }$ for long sticks. So, $S1$ can arrange with any $\text{L1,L2,....,L10 }$, i.e. 10 ways. Then $S2$ can arrange with any $\text{L1,L2,....,L9}$,i.e. 9 ways etc. So, the total combination of short and long is 10!
So, the probability = $\frac{10!}{{}_{20} C_{2}}$
But this is not the right answer. Can you please someone give me the right answer & the logic behind the answer. Also, if someone can explain why my answer is wrong.
Thanks!
 A: For pairing $2n$ items, the formula is $\dfrac{(2n)!}{n!2^n}$ which can be understood by noting that in the $(2n)!$ permutations of the items which we can pair serially, neither the order of the pairs nor the order within the pairs matter, thus the division by $n!2^n$
I  much prefer, though, to think of the ways for sequentially forming pairs, ie $[(2n-1)(2n-3)(2n-5)...1]$
This skipping by two is called a double factorial
and  would be denoted as $(2n-1)!!$
Thus the probability of pairs being formed has a very simple formula,
$Pr = \dfrac{n!}{(2n-1)!!}$, here yielding $\dfrac{10!}{19!!} \approx 0.0055$
A: Alternative approach:
Assume that the $(20)$ items will be laid out in a row.  There are
$$(20)!$$
ways of doing this.  Then, the question is, how many satisfying ways are there, of laying out the items.
You can presume that items $(2k-1),(2k)$ will be paired together, for $k \in \{1,2,\cdots,10\}$.
Think of the pairs of items as pair-units.  So, each short item must be assigned to one of the pair-units.  There are $(10)!$ ways of doing this.
Similarly, there are $(10)!$ ways of assigning the long items to the $(10)$ pair-units.
Further, within each pair-unit, the short and long items can be permuted in $(2!) = 2$ ways.
Therefore, the number of satisfying permutations is
$$2^{10} \times [(10)!]^2.$$
Therefore, the overall probability is
$$\frac{2^{10} \times [(10)!]^2}{(20)!}.$$
