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  4. The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by a!× b! , then (a/b) is

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Q. The number of ways in which $6$ men and $5$ women can dine at a round table if no two women are to sit together is given by $a!\times b! $, then $\frac{a}{b}$ is

Permutations and Combinations

Solution:

No. of ways in which $6$ men can be arranged at a round table $=(6-1)!=5!$
Now women can be arranged in $^{6}P_{5}$
$=6!$ ways
Total Number of ways $=6!\times 5!= a! \times b!$
$\Rightarrow \frac{a}{b}=\frac{6}{5}=1.2$