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Difficult Math Question #26 - Combinations

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by 800guy » Mon Oct 09, 2006 4:34 pm
OA coming after some people answer

Seven men and seven women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?
a.24 b.6 c.4 d.12 e.3
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by ajith » Thu Oct 19, 2006 12:04 am
One approach is like this

arrangee 7 men around the table that can be done in 6! ways..
and arrange women in the gaps between men
there are 7 gaps.. and 7 women so it can be done in 7! ways ..

so total no: of arrangements is seeming to be 6!*7!...

Which is not in the options..

Are you sure that the number of men and women are correct...
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OA

by 800guy » Fri Oct 20, 2006 10:34 am
here's the OA:

I suggest to first arranging men. This can be done in 6! Ways. Now to satisfy above condition for women, they should sit in spaces between each man. This can be done in 7! Ways (because there will be seven spaces between each man on round table)
Total ways = 6! * 7!
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Re: OA

by veekay » Sat Aug 25, 2007 10:28 am
800guy wrote:here's the OA:

I suggest to first arranging men. This can be done in 6! Ways. Now to satisfy above condition for women, they should sit in spaces between each man. This can be done in 7! Ways (because there will be seven spaces between each man on round table)
Total ways = 6! * 7!
I know this is an old thread. Appreciate if anyone could explain....

Why will it be 6! 7!. I think it should be 7!7! as there are 7 places for men and men can be arranged in 7! ways
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by beny » Tue Aug 28, 2007 2:03 am
The table is circular... look into rules for circular permutations.
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