Question

The number of ways in which $200$ different things can be divided into groups of $100$ pairs is:

• A. $(1.3.5.....199)$
• B. $\left(\frac{101}{2}\right)\left(\frac{102}{2}\right)\left(\frac{103}{2}\right)....\left(\frac{200}{2}\right)$
• C. $\frac{200!}{2^{100}\left(100\right)!}$
• D. $Allofthese$

Answer 1

The correct option is C $\frac{200!}{2^{100}\left(100\right)!}$

We need to form $100$ groups having $2$ objects each.

Number of possible ways $=200!$

Now, we eliminate intrapermutation $200!$ by dividing $2\times 2\times \mathrm{2......}$ upto $100=2^{100}$
200by
=210so as to account for the intrapermutations of each group.

Since, the group size is the same,the $200!$200! we initially got accounts for those cases also in which we are permuting the $100$100 groups themselves in $100!$ ways.

$=\frac{200!}{2^{100}\times 100!}$