Question

The number of ways in which $100$ people can be divided in $50$ pairs is

• A. $\frac{100!}{2^{50}(50!)}$
• B. $\frac{100!}{2^{50}}$
• C. $\prod _{n=26}^{50}\frac{{}^{2n}{C}_2}{2}$
• D. $\prod _{n=1}^{50}{}^{2n}{C}_2$

Answer 1

Answer: (A), (C)

$\prod _{n=26}^{50}\frac{{}^{2n}{C}_2}{2}$

Number of ways of arranging $100$ people is $100!$
But, when we make a pair then both the persons in a pair are considered as one person. So, we divide it by $2!\times 2!\times 2!\cdots \text{upto}50$ times.
So, we get $\frac{100!}{2^{50}}$ ways.
But, the above case considers the arrangement of $50$ couples also which should be eliminated.
So, the required number of ways is $\frac{100!}{2^{50}\times 50!}$

Also,

$\frac{100!}{2^{50}\times 50!}=\frac{51}{2}\times \frac{52}{2}\times \frac{53}{2}\times ...\times \frac{100}{2}$
$=\prod _{n=26}^{50}\frac{{}^{2n}{C}_2}{2}$