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Question

n(≥5) persons are sitting in a row. Three of these are selected at random, the probability that no two of the selected persons are together is

A
n3P2nP2
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B
n3C2nC2
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C
(n3)(n4)n(n1)
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D
n3C2nP2
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Solution

The correct options are A n−3C2nC2 B (n−3)(n−4)n(n−1) C n−3P2nP2Select 3 persons (resulting in 2 gaps in between them and 2 in the extremes i.e. 4 gaps in all). Place 1 person each in these gaps so that the 3 persons are not consecutive. The remaining (n−5) persons can be distributed in the 4 gaps in Cn−5+4−14−1=Cn−23=(n−2)(n−3)(n−4)6 ways. Hence, probability = Cn−23Cn3=(n−2)(n−3)(n−4)6n(n−1)(n−2)6=(n−3)(n−4)n(n−1)

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