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Question

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

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Solution

Out of 21, we can choose 15 for one table in 21C15 ways
And for each selection we are left with 6 guests for the second table having 6 seats.
This can be chosen in 6C6 ways.
Now 15 for round table A can be arranged in (14)! ways and 6 for round table B can be arranged in 5! ways.
Hence the total number is 21C15×6C6×14!×5!.

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