UPSC NDA II 2023 Mathematics Solved Paper

Concept:Use the gap method to ensure no two girls sit together.Explanation:First, arrange the three boys in a row. This can be done in $3! = 6$ ways.After placing the boys, there are 4 gaps (before the first boy, between boys, after the last boy).To keep no two girls together, each girl must occupy a distinct gap.Choose 3 out of 4 gaps for the girls and arrange the three girls in those chosen gaps. This can be done in ${}^4P_3 = 4 \times 3 \times 2 = 24$ ways.So total favourable arrangements = $6 \times 24 = 144$.Total possible arrangements of 6 distinct persons = $6! = 720$.Required probability = $\frac{144}{720} = \frac{1}{5}$.Answer:$\frac{1}{5}$ (Option D).