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

Answer: 1.2

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 $5!\times 6!\Rightarrow n=5!\times 6!$
$\Rightarrow k=\frac{m}{10n}=\frac{2\times 6!\times 6!}{10\times 5!\times 6!}=\frac{12}{10}=1.2$