Question

A total of $6$ Boys and $6$ girls are to sit in a row alternatively and in a circle. Let $\text{m}$ be the number of arrangements in the row and $\text{n}$ be the number of arrangements in the circle. If $\text{k}=\frac{\text{m}}{10 \text{n}}$ , then the value of $\text{k}$ is

Answer 1

Linear permutations with boy in the first place are of the form $B_{1}G_{1}B_{2}G_{2}B_{3}G_{3}B_{4}G_{4}B_{5}G_{5}B_{6}G_{6}$ and the number of such arrangements is $6!\times 6!.$
The number of linear permutations with girl in the first place is $6!\times 6!$
So, the number of arrangements in row $=m=2\times 6!\times 6!$
(In circular permutations, start with a place which can be filled by a boy or a girl and after that the arrangements become linear)
Placing the boys first and then arranging the girls in $6$ gaps, the number of such circular arrangements is $5!\times 6!\Rightarrow n=5!\times 6!$
$\Rightarrow k=\frac{m}{10 n}=\frac{2 \times 6 ! \times 6 !}{10 \times 5 ! \times 6 !}=\frac{12}{10}=1.2$