Question

$30$ persons were invited for a party. In how many ways they and a host can be seated round the table so that two particular person always remain on both side of the host?

• A. $(31!\times 2!)$
• B. $\left(30!\times 3!\right)$
• C. $(30!\times 2!)$
• D. $(28!\times 2!)$

Answer 1

The correct option is D $(28!\times 2!)$

30 persons + host = 31 persons
$\frac{P_{1}H{P}_2}{}+28\mathrm{persons}H\to \mathrm{host}{P}_1\mathrm{and}{P}_2\mathrm{are}\mathrm{two}\mathrm{particular}\mathrm{persons}\mathrm{sorrounding}\mathrm{host}$
Here, we have to arrange $P_1{\mathrm{HP}}_2$ (consider this group as one person) and 28 persons in circle.
the possible permutations$(29-1)!=28!$
Now,$P_1\mathrm{and}{P}_2\mathrm{can}\mathrm{arrange}\mathrm{in}\mathrm{two}\mathrm{ways}.$ so, total number of permutations = $28!2!$
answer is option(d)