Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. $20$ persons are invited for a party In how many different ways can they and the host be seated at circular table, if the two particular persons are to be seated on either side of the host?

Permutations and Combinations

Solution:

There are total $20 + 1 = 21$ persons. The two particular persons and the host be taken as one unit so that these remain $21 - 3 + 1 = 19$ persons be arranged in round table in $18!$ ways. But the two persons on either sides of the host can themselves be arranged in $2!$ ways.
$\therefore $ Required number of ways $= 2! \times 18! = 2 \times 18! $