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.

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 $$12! - 7!6!$$

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

• Reminds me of math.stackexchange.com/questions/697433/… – lab bhattacharjee Oct 25 '18 at 12:47
• For no 2 girls can sit together - If we arrange Boys and Girls alternatively we can arrange them in $6! * 6!$ ways. But we can also switch them to form a new arrangement. Hence, $6!*6!*2$. Is this correct? – Kaushik Aug 6 at 21:21

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.