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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?

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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

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  • $\begingroup$ Reminds me of math.stackexchange.com/questions/697433/… $\endgroup$ – lab bhattacharjee Oct 25 '18 at 12:47
  • $\begingroup$ 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? $\endgroup$ – Kaushik Aug 6 at 21:21
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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.

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