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There are 18 guests at a dinner party. They have to sit 9 guests on either side of a long table, three particular persons decide to sit on one side and two others on the other side. In how many ways can the guests to be seated?

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#### Solution

Let A and B be two sides of the table 9 guests sit on either side of the table in 9! × 9! ways.

Out of 18 guests, three particular persons decide to sit namely inside A and two on the other side B. remaining guest = 18 – 3 – 2 = 13.

From 13 guests we can select 6 more guests for side A and 7 for the side.

Selecting 6 guests from 13 can be done in ^{13}C_{6} ways.

Therefore total number of ways the guest to be seated = ^{13}C_{6} × 9! × 9!

`= (13!)/(6!(13 - 6)!) xx 9! xx 9!`

`= (13!)/(6! xx 7!) xx 9! xx 9!`

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