A closet contains $13$ different pairs of gloves. The 26 gloves are randomly arranged into $13$ pairs.
(a) Find the probability that all left-hand gloves are paired with right-hand gloves (not necessarily matching).
(b) Find the probability that all of the gloves are arranged so that each left-hand glove is paired with its matching right-hand glove.
Updated:
To deal with (a). First I get the number of ways to arrange 26 gloves into pairs ($n=13$), the order does not matter: $$\frac{(2n)!}{2^{n}}\cdot\frac{1}{n!}$$
After that, I calculate the number of ways in which all left-hand gloves are paired with right-hand gloves (not fully matching): $n!=13!$
So the probability of (a) equals: $$n!\cdot\frac{(2^{n})n!}{(2n)!}$$
And the probability of (b) equals (only one case when all gloves are fully matching): $$\frac{(2^{n})n!}{(2n)!}$$
Is my approach correct?
Best Answer
Your solutions are correct.
There is another way to count the number of elements in the sample space. Suppose we have $n$ pairs of gloves in the closet. Grab any of the $2n$ gloves in the closet. It does not matter which glove we choose. There are $2n - 1$ gloves left in the closet. Pair one of them with the first glove you select. Set that pair aside. Take any of the remaining $2n - 2$ gloves out of the closet. Again, it does which glove we choose. Pair any of the remaining $2n - 3$ gloves left in the closet with that glove. Iterate the process. The number of ways of forming $n$ pairs of gloves is $$(2n - 1)!! = (2n - 1)(2n - 3)(2n - 5) \cdots (5)(3)(1)$$ where $k!!$ is referred to as the double factorial.
We can verify that this expression is equivalent to your expression as follows. Based partly on work you did in the comments, you found that the number of ways to place the gloves in pairs is $$\frac{1}{n!2^n}\binom{2n}{2}\binom{2n - 2}{2}\binom{2n - 4}{2} \cdots \binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{(2n)!}{n!2^n}$$ Observe that \begin{align*} \frac{(2n)!}{n!2^n} & = \frac{(2n)(2n - 1)(2n - 2)(2n - 3) \cdots (4)(3)(2)(1)}{n(n - 1)(n - 2)(n - 3) \cdots (3)(2)(1)(2^n)}\\ & = \frac{(2n - 1)(2n - 3)(2n - 5) \cdots (5)(3)(1)(n!)}{n!}\\ & = (2n - 1)(2n - 3)(2n - 5) \cdots 5 \cdot 3 \cdot 1\\ & = (2n - 1)!! \end{align*} where we first cancel a $2$ from each even factor to obtain the expression in the second line, then cancel $n!$ to obtain the expression in the third line.
In this particular problem, $n = 13$, so the number of elements in our sample space is $$25!! = 25 \cdot 23 \cdot 21 \cdot 19 \cdot 17 \cdot 15 \cdot 13 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$$
If we place the right-hand gloves in a line (in any order), there are $n$ left-hand gloves which can be matched with the first right-hand glove, $n - 1$ left-hand gloves which can be matched with the second right-hand glove, $n - 2$ left-hand gloves which can be matched with the third right-hand glove, and so forth. Hence, there are $n!$ favorable cases. Thus, the probability that each left-hand glove is matched with a right-hand glove is $$\frac{n!}{(2n - 1)!!}$$ In this particular problem, we obtain $$\frac{13!}{25!!} = \frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{25 \cdot 23 \cdot 21 \cdot 19 \cdot 17 \cdot 15 \cdot 13 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1}$$
There is only one favorable case. Hence, the probability is $$\frac{1}{(2n - 1)!!}$$ In this particular problem, we obtain $$\frac{1}{25 \cdot 23 \cdot 21 \cdot 19 \cdot 17 \cdot 15 \cdot 13 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1}$$