Question

$12$ boys and $2$ girls are to be seated in a row such that there are at least $3$ boys between the $2$ girls. The number of ways this can be done is $m\times 12!$ where $m=?$

• A. $2\times {}^{12}C_6$
• B. $20$
• C. ${}^{11}P_2$
• D. ${}^{11}C_2$

Answer 1

The correct option is C

${}^{11}P_2$

Explanation for the correct option:

Find the value of $m$:

Given that $12$ boys and $2$ girls are to be seated

So, arrangement of two girls sits together $=2x13$

$=26$

arrangement of one boy sits between two girls $=2x12$

$=24$

arrangement of two boys sits between two girls $=2x11$

$=22$

Total seating arrangement $=14!$

Thus, Required arrangement $=14!-(26+24+22)12!$

$={}^{11}P_2\times 12!$

$\therefore$ $\mathit{m}\mathbf{=}{}^{\mathbf{11}}\mathit{P}_{\mathbf{2}}$

Hence, Option ‘C’ is Correct.