Find the number of ways in which 6 boys and 6 girls can be seated in a row so that all the girls are never together.
Attempt:
total number of ways – number of ways in which all girls are together
$= 12! – 2!\times(6!)\times (6!)$
But answer given is $12! – 7!6!$
Is the given answer wrong?
Best Answer
With any constraint the number of possible combination is $12!$
If all the girls are together, we can think the set to be of $6+1$ members which can be arranged in $7!$ ways
and again the $6$ girls can be arranged in $6!$ ways