how to make pairs from odd number of people one may be alone?

Question

How Many ways the pairs can be formed from the group of size odd(like 3,5), one may be left alone?

Eg: there are 5 students, so 2 pairs can be formed and one guy left alone...

|Consider (a,b,c,d,e) are 5 distinct person , According to me only 6 ways are possible

|  (a,b)  |     (a,c)       |   (a,d)         |      (a,e)        |
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|  (c,d)  |  (b,e)          | (b,c)           |  (b,c)(redundant) |
|  (c,e)  |  (b,d)          | (b,e)(redundant)|  (b,d)(redundant) |
|  (d,e)  | (d,e)(redundant)| (c,e)(redundant)|  (c,d)(redundant) |
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Is it correct? what will be the formula for this?

Answer 1

If there are $n$ people you can choose the first pair in $n \choose 2$ ways, then the next in $n-2 \choose 2$ ways, and so on. You have made $\frac 12(n-1)$ groups and you could have made the groups in any order, so the number of ways is $${n \choose 2}{n-2 \choose 2}\ldots {2\choose 2}\frac 1{(\frac 12(n-1))!}=\frac{n!}{2^{(\frac 12(n-1))}}\frac 1{(\frac 12(n-1))!}$$ For $n=5$ this evaluates to $15$. There are $10$ ways to pick the first pair, $3$ ways to pick the second, and you divide by $2$ for the orders to pick the pairs. There are the twelve cases you list in your table (including the redundant ones) plus three where $a$ is alone.

Answer 2

You have to count the number of possible configurations $$ \{*\},\{*,*\},\{*,*\} $$ You can choose the single element in $5$ ways, then the first pair in $\binom{4}{2}$ ways. However, you have to divide by $2!$, because you can permutate the pairs: $$ 5\cdot\binom{4}{2}\cdot\frac{1}{2!}=5\cdot6/2=15 $$ There would also be a $\binom{2}{2}$ factor for the last pair, which I omitted.

Similarly for $2n+1$ elements: $$ (2n+1)\binom{2n}{2}\binom{2n-2}{2}\dotsm\binom{4}{2}\binom{2}{2}\cdot\frac{1}{n!} $$ Can we do a recursive formula? Yes, if $P_n$ is the number corresponding to $2n+1$ elements, then $P_0=1$ and $$ \frac{P_{n+1}}{P_n}=\frac{2n+3}{2n+1}\binom{2n+2}{2}\frac{1}{n+1}=2n+3 $$ Indeed, $$ P_1=3,\quad P_2=(2\cdot1+3)\cdot3=15 $$ and so on.

Since $P_{n+1}=(2(n+1)+1)P_n$, if we set $Q_{2n+1}=P_n$, we have $Q_k=k!!$ (the semifactorial), for odd $k$.