Probability Question 431
Question: 5 boys and 5 girls are sitting in a row randomly. The probability that boys and girls sit alternatively is
[Kerala (Engg.) 2005]
Options:
A) 5/126
B) 1/126
C) 4/126
D) 6/125
Show Answer
Answer:
Correct Answer: B
Solution:
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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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