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.

Question

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?

score 4 · Accepted Answer · answered Oct 25 '18 at 12:09

4

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

answered Oct 25 '18 at 12:09

lab bhattacharjee

Reminds me of https://math.stackexchange.com/questions/697433/a-family-has-three-children-what-is-the-probability-that-at-least-one-of-them-i/697436#697436 – 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 06 '19 at 21:21

score 2 · Answer 2 · answered Oct 25 '18 at 12:10

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.

2 Answers2