Question

If $20$ men are to be seated on a long table having $10$ chairs on either side. $3$ particular men want to sit on one particular side and $5$ particular men want to sit on other side, then number of possible arrangement is

• A. $12!\cdot {}^{10}{P}_3\cdot {}^{10}{P}_5$
• B. $2\cdot {}^{10}{P}_3\cdot {}^{10}{P}_5$
• C. $2\cdot 12!\cdot {}^{10}{P}_3\cdot {}^{10}{P}_5$
• D. ${}^{10}{P}_3\cdot {}^{10}{P}_5$

Answer 1

The correct option is A $12!\cdot {}^{10}{P}_3\cdot {}^{10}{P}_5$

Let $a_1,a_2,\cdots ,a_{10}$ and $b_1,b_2,\cdots ,b_{10}$ be the chairs on each side.
Now,
$3$ particular men sit on one particular side on $10$ chairs in $=\left({}^{10}{P}_3\right)$ ways and
$5$ particular men sit another side on $10$ chairs $={}^{10}{P}_5$ and
Rest men can be seated in $12!$ ways
$\therefore$Total possible arrangement $=12!\cdot {}^{10}{P}_3\cdot {}^{10}{P}_5$