Let $m$ be the number of ways of arranging 5 boys and 5 girls in a row such that boys and girls sit alternately. Let $n$ be the number of ways of arranging 5 boys and 5 girls in a row such that no two girls sit together then $\frac{n}{m}$ equals

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Detailed Solution

Boys can be arranged in 5! Ways

$X B_{1} X B_{2} X B_{3} X B_{4} X B_{5} X$

Girls can be arranged in first 5 gaps or last 5 gaps in 2.5!

$∴ m = 2.5! 5!$

Now 5 girls can be arranged in 6 gaps in $^{6} P_{5} = 6!$ ways

$∴ n = 5! 6!$

Now, $\frac{n}{m} = \frac{5! 6!}{2.5! 5!} = 3$ .

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