Arranging persons so that any of the ladies and childern does not sit together

Question

In what ways 2 Men and 3 ladies and 2 childern can be sit so that any of ladies and any of childern do not sit together?(however any childern and any lady can sit together)

I have solved above question with a different method but I am confused how to solve this with gap method or slot method

Please help

I will take the time to restate your question in a better way. — Yash Jain, Jan 25 '18 at 12:34
Where are these people sitting? Around a circular table? Along a line with two ends? In several rows or at several tables? Are there only 7 chairs or enough to leave some chairs empty? — hmakholm left over Monica, Jan 25 '18 at 13:40
actually my question is edited. my actual question was any woman should not sit together and any child should not sit together — Neeraj Sharma, Jan 26 '18 at 04:35
Now, @Neeraj The question is back to as you wrote it. Please edit the question to make clear what you mean. — amWhy, Jan 27 '18 at 01:14

true blue anil · Answer 1 · 2018-01-25T15:46:01.040

All problems can't necessarily be straight-jacketed into a particular mould, here's one way.

The men and children can be placed in $4! = 24$ ways,
of which, by symmetry, $12$ will be with the children together, and $12$ with the children apart.

If the children are together, eg $\bullet MCCM\bullet\;\; or\;\; \bullet M\bullet MCC$,
$2$ gaps will be available for placing the first woman, and gaps will increase with each placement, so the women can be placed in $2\cdot3\cdot4 = 24$ ways.

And if the children are apart, eg$\;CM\bullet MC$ or $\bullet MCMC$,
only $1$ gap will be available for the first woman, so $1\cdot2\cdot3 = 6$ ways

Putting the pieces together, answer = $12(24+6) = 360$ ways

Arranging persons so that any of the ladies and childern does not sit together

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