A) \[\frac{8}{15}\]
B) \[\frac{4}{15}\]
C) \[\frac{2}{15}\]
D) \[\frac{1}{15}\]
Correct Answer: B
Solution :
Consider a group of three students A, B and an other student in between A and B. Choice for a student between A and B is and B can interchange their places in the group in 2 ways. Now the group of three students (student A, student B and a student in between A and B) and the remaining 3 students can be stand in a row in 4! ways. Hence total number of ways to stand in a row such that A and B are separated with one student in between them \[=4\times 2\times 4!\] Now total number of ways to stand 6 student stand in a row without any restriction = 6! Hence required probability \[=\frac{4\times 2\times 4!}{6!}=\frac{4\times 2}{6\times 5}=\frac{4}{15}\]
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