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?
On
All the girls together means a sequence of 6 girls in a row. This sequence can start at position 1, 2, 3, 4, 5, 6 and 7.
There are $6!$ ways to place the girls and $6!$ ways to place the boys.
So you have $7 \cdot 6! \cdot 6! = 7! \cdot 6!$ possible ways to violate the rule.
Thus there are $12! - 7! \cdot 6!$ ways that respect the rule.
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