A) \[\frac{5}{126}\]
B) \[\frac{1}{42}\]
C) \[\frac{4}{126}\]
D) \[\frac{6}{126}\]
E) \[\frac{1}{63}\]
Correct Answer: B
Solution :
Total number of ways\[=10!\] Total number of ways in which 5 boys and 5 girls are sitting in a row randomly \[=5!\,\times 6!\] \[\therefore \] Required probability\[=\frac{5!\,\times 6!}{10!}=\frac{1}{42}\]
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