A) \[\frac{1}{34}\]
B) \[\frac{1}{35}\]
C) \[\frac{1}{17}\]
D) \[\frac{1}{68}\]
Correct Answer: B
Solution :
Four boys can be arranged in \[4\,!\] ways and three girls can be arranged in 3! ways. \[\therefore \] The favourable cases \[=4\,!\,\times \,3\,!\] Hence the required probability \[\frac{=4\,!\,\times 3\,!}{7\,!}=\frac{6}{7\times 6\times 5}=\frac{1}{35}\].
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