Question

A person invites a group of $10$ friends to dinner and makes them sit on $2$ round tables with $5$ persons at each table. The number of ways in which he can arrange the guests is:

• A. $\frac{10!}{5}$
• B. $\frac{10!}{25}$
• C. $\frac{10!}{5!}$
• D. $\frac{10!}{50}$

Answer 1

The correct option is B $\frac{10!}{25}$

The number of ways of selection of $5$ friends for first table can be done in ${}^{10}{C}_5$ ways.
Remaining $5$ friends can be arranged on second table.

The total number of permutations of $5$ guests on each table is $4!$
Hence, the total number of arrangements
$={}^{10}{C}_5\times 4!\times 4!=\frac{10!}{(5!\times 5!)}\times 4!\times 4!=\frac{10!}{25}$