Algebra - Elements of Combinatorics

Five boys and three girls are sitting in a row of eight seats. In how many ways can they be seated so that not all the girls sit side by side?


Consider the three girls as $1$ group.

Then there are $5$ boys AND $1$ group of three girls, i.e. $6$ people.

These can be seated amongst themselves in $6! = 720$ ways.

Amongst themselves, the $3$ girls can be seated in $3! = 6$ ways

$\therefore \;$ Number of ways in which $5$ boys AND $1$ group of three girls can be seated

$= 3! \times 6! = 6 \times 720 = 4320$ ways.

Now, $3$ girls and $5$ boys, i.e. $8$ people can be seated amongst themselves in $8! = 40320$ ways.

$\therefore \;$ Number of ways in which $3$ girls and $5$ boys can be seated so that not all the girls sit next to one another

$= 40320 - 4320 = 36000$ ways