Question

A person invites a group of 10 friends at dinner and sits 5 on a round table and 5 more on another round table, 4 on one round table and 6 on the other round table. Find the number of ways in each case in which he can arrange the guest.

Answer 1

(i) The number of ways of selection of 5 friends for the first table is `.^(10)C_(5)`. Remaining 5 friends are left for the second table.
The total number of permutations of 5 guests on each table is 4!. Hence, the total number of arrangements is
`.^(10)C_(5)xx4!xx4!=((10)!)/(5!xx5!)xx4!xx4!=10!//25`.
(ii) The number of ways of selection of 6 guests is `.^(10)C_(6)` The number of ways of permutations of 6 guests on round table is 5!. The number of permutation of 4 guests on another round table is 3!.
Therefore, total number of arrangements is
`.^(10)C_(6)xx5!xx3!=((10)!)/(6!xx4!)5!xx3!=((10)!)/(24)`