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?

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

No it is not correct. The correct answer is $\displaystyle 2!\cdot 3!$

You tell us why.

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...

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

it's tricky, but how did you get that 2! * 3! so fast?

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...

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

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 $\displaystyle \frac{6!}{6}=5!$.

There are $\displaystyle \frac{3!}{3}=2!$ ways to seat the men in every other chair.

There are $\displaystyle 3!$ ways to seat the women at the remaining, now ordered, table.