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8 years, 4 months ago
8 years, 4 months ago
Suppose there are 7 seats numbered 1,2,3,4,5,6 and 7. Now seat no. 4 will always be occupied and no one can take seats 3 and 5 as there can not be two person sitting on consecutive seats. So we can sit one person in seat 1 or 2 and the other in seat 6 or 7. thus total number of ways in which 3 person sit according to the given condition will be 2*2 = 4.
So required probability = 4/(7C3)=4/35.
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