A) \[\frac{1}{3}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) \[\frac{2}{3}\]
Correct Answer: A
Solution :
[a] Seven people can seat themselves at a round table in 6! Ways. The number of ways in which two distinguished persons will be next to each other \[r=2(5)!\]. Hence, the required probability \[=\frac{2(5)!}{6!}=\frac{1}{3}.\]
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