A) 5/126
B) 1/126
C) 4/126
D) 6/125
E) 1/63
Correct Answer: B
Solution :
Let n = total no. of ways = 10! m = favourable no. of ways = 2 × 5! . 5! Since the boys and girls can sit alternately in 5 ! . 5! ways if we begin with a boy and similarly they can sit alternately in 5! . 5! ways if we begin with a girl Hence, required probability = \[\frac{m}{n}\] = \[\frac{2\times 5!.5!}{10!}\]= \[\frac{2\times 5!}{10\times 9\times 8\times 7\times 6}\] = \[\frac{1}{126}\].
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