Question

$12$ boys and $2$ girls are to be seated in a row such that there are atleast $3$ boys between the $2$ girls. The number of ways this can be done is $\lambda \times 12!$, then the value of $\lambda$, is

• A. $55$
• B. $110$
• C. $20$
• D. $45$

Answer 1

The correct option is B $110$

arrangement in two girls are together = 2 x 13 = 26

=> one boy sit between two girls = 2 x 12 = 24

=> two boys sit between two girls = 2 x 11 = 22

=> Total arrangements = 14!

$\therefore$ Required arrangement $14!-(26+24+22)12!{=}^{11}{\text{P}}_{2}12$

$\therefore \lambda {=}^{11}{\text{P}}_2=110$

Answer