Six people permutation

• Apr 17th 2012, 01:05 PM
Nforce
Six people permutation
We have 6 people, 3 men, and 3 women.

On how many ways, can we sat them at a round table, on 6 chairs, but they must sit in this order: man, woman,man,woman... So the neighbour must be the opposite sex.

Is this correct?

$\frac{3!*3!}{6}$

the six in the denominator is because it's a round table, and we can rotate them for one place, and would be the same so it's 6 for six places.
• Apr 17th 2012, 01:35 PM
Plato
Re: Six people permutation
Quote:

Originally Posted by Nforce
We have 6 people, 3 men, and 3 women.
On how many ways, can we sat them at a round table, on 6 chairs, but they must sit in this order: man, woman,man,woman... So the neighbour must be the opposite sex.
Is this correct?
$\frac{3!*3!}{6}$

No it is not correct. The correct answer is $2!\cdot 3!$
You tell us why.
• Apr 17th 2012, 01:53 PM
Nforce
Re: Six people permutation
Ok, I made this on a long way, first man has 6 choices, second 2 and third one, and for the woman 3!, and because it's round we devided by 6.
and, get...

$\frac{6*2*1*3!}{6} = 12$

it's tricky, but how did you get that 2! * 3! so fast?
• Apr 17th 2012, 02:09 PM
Plato
Re: Six people permutation
Quote:

Originally Posted by Nforce
Ok, I made this on a long way, first man has 6 choices, second 2 and third one, and for the woman 3!, and because it's round we devided by 6.
and, get...
$\frac{6*2*1*3!}{6} = 12$
it's tricky, but how did you get that 2! * 3! so fast?

The way you did it works. It is just different.
To seat the six at the table in any order is $\frac{6!}{6}=5!$.
There are $\frac{3!}{3}=2!$ ways to seat the men in every other chair.
There are $3!$ ways to seat the women at the remaining, now ordered, table.