Question

The total number of ways in which $3$ girls and $3$ boys be seated at a round table, so that any $2$ and only $2$ of the girls are always together is

Answer 1

$3$ boys can be arranged in circular manner is $=\frac{3!}{3}=2!$ ways


Now, two girls (always seated together ) can be selected from 3 girls in ${}^3{C}_2$ ways

Selecting one gap from circular arrangement of Boys can be done in ${}^3{C}_1$ ways.

Arrangement of two girls to be seated together in the gaps between Boys can be done in
$={}^3{C}_1.{}^3{C}_2.2!$ (as girls can arrange themselves)

Now, one more girl can be arrange in remaining $2$ places in $2!$ ways
$\therefore$ Required number of arrangements $=2!({}^3{C}_1.{}^3{C}_2.2!)2!=72$ ways