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 in $2!$ 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.
The number of ways two girls can be seated together in the gaps between boys
$={}^3{C}_1\cdot {}^3{C}_2\cdot 2!$
(as girls can arrange themselves)

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