A) \[\frac{1}{35}\]
B) \[\frac{1}{14}\]
C) \[\frac{1}{15}\]
D) None of these
Correct Answer: B
Solution :
Total ways of arrangements \[=\frac{8\,!}{2\,!\,\,.\,\,4\,!}\] \[\bullet \,w\bullet x\bullet y\bullet z\bullet \] Now ?S? can have places at dot?s and in places of \[w,\,\,x,\,\,y,\,\,z\] we have to put \[2\,A's,\] one \[I\] and one \[N.\] Therefore favourable ways \[=5\text{ }\left( \frac{4\,!}{2\,!} \right)\] Hence required probability \[=\frac{5\,.\,4\,!\,\,2\,!\,\,4\,!}{2\,!\,\,8\,!}=\frac{1}{14}.\]
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