A businessman hosts a dinner to 21 guests. He has 2 round tables which can accommodate 15 and 6 persons each. In how many ways can he arrange the guests?

Given that we need to accommodate 21 guests to two round tables which can accommodate 15 and 6 persons each.

We need to select 6 members first and arrange 6 and 15 members accordingly in the respective tables.

Let us assume the no. of ways of choosing 6 members to be N1

N1 = No. of ways of choosing 6 members out of 21 guests.

N1 = 21C6

Now we need to arrange the 6 members in a round table. By fixing a guest at a single seat, We arrange the remaining 5 members. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of arrangements that can be made is 5!.

Now we need to arrange the 15 members in a round table. By fixing a guest at a single seat, We arrange the remaining 5 members. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of arrangements that can be made is 14!.

Let us assume total no. of ways of arranging the guests in the table be N

N = N1 × 5! × 14!

N = 21C6 × 5! × 14!

The no. of ways of accommodating guests is 21C6 × 5! × 14!.

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