Question

A person invites a party of $10$ friends at dinner and placed of them are
(i) $5$ at one round table, $5$ at the other round table
(i) $4$ at one round table, $6$ at the other round table
Then the ratio of number of cirular permutations of case
(i) and case (ii) is

• A. $24/13$
• B. $25/24$
• C. $24/25$
• D. $13/24$

Answer 1

The correct option is C $24/25$

$\Rightarrow$ He selected $5$ persons among $10$ in ${}^{10}{C}_5$ ways.

$\Rightarrow$ Circular permutations of $5$ guests in one table $=(5-1)!ways=4!$

$\Rightarrow$ Similarly for $2nd$ table number of circular permutation $=4!$

$\therefore$ Total number of arrangements ${=}^{10}{C}_5\times 4!\times 4!=\frac{10!}{125}$

$\Rightarrow$ Again number of selectors of $6$ guests for first table ${}^{10}{C}_6$

$\Rightarrow$ Number of permutations of $6$ guests on one round table $=(6-1)!=5!$

$\Rightarrow$ And number of permutations of $4$ guets on second table $(4-1)!=3!$

$\therefore$ Total number of arrangements ${=}^{10}{C}_5.5!3!=\frac{10!}{24}$

$\therefore$ Required ratio $=\frac{10!/25}{10!/24}=\frac{24}{25}$