Browse Questions

# 6 boys and 6 girls sit in a row at random.The probability that all the girls sit together is

$\begin{array}{1 1}(A)\;\large\frac{1}{432}\\(B)\;\large\frac{12}{432}\\(C)\;\large\frac{1}{132}\\(D)\;\text{None of these}\end{array}$

Step 1:
Given 6 boys and 6 girls
$\therefore$ Number of ways in which 6 boys and 6 girls together sitting in a row =7!
6 girls sitting arrangement =6!
Step 2:
$\therefore$ Required probability =$\large\frac{7!\times 6!}{12!}$
$\Rightarrow \large\frac{1}{132}$
Hence (C) is the correct answer.