Permutation Formula Question

Question

The problem that I am having trouble figuring out is:

In how many ways can give five men and five girls be seated at a round table so that each girl is between two men?

I know that the formula for permutations is:

The number of permutations of n distinct objects taken r at a time is $P(n,r) = \frac {n!}{(n-r)!}$

Do I just substitute into the formula so that n = 10 (girls + men) and r = 2? Or is there an easier way to do this without the formula? Thanks in advance for any help.

Answer 1

If the chairs are fixed and cannot be moved, then there are two possibilities for where men and where women will sit. In each of them, you have 5! possibilities for men and 5! for women, so in total $2 \times (5!)^2$. However, if you don't distinguish rotations and consider $(a_1,\ldots, a_{10})$ to be the same as $(a_2,\ldots, a_{10}, a_1)$, then it's only $4!\times 5!$, because a fixed man sits "somewhere" and you have $4!$ options for the remaining 4 men; and 5! permutations for the women.

Answer 2

Label the seats on the table $1,2,3\dots 10$ Then there are two possibilities when we look at the people under seat 1,2,3 etc. :

$\underbrace{G}_1\underbrace{B}_2\underbrace{G}_3\underbrace{B}_4\underbrace{G}_5\underbrace{B}_6\underbrace{G}_7\underbrace{B}_8\underbrace{G}_9\underbrace{B}_{10}$

$\underbrace{B}_1\underbrace{G}_2\underbrace{B}_3\underbrace{G}_4\underbrace{B}_5\underbrace{G}_6\underbrace{B}_7\underbrace{G}_8\underbrace{B}_{9}\underbrace{G}_{10}$

But notice every one of the second combinations can be achieved by asking everyone to scoot over 1 seat, every one of the first combinations generates $9$ others when we ask the people to scoot over $1$,$2\dots9$ places.

Therefore what we want is the number of combinations of those forms
divided by 10.

There are $5!$ ways to pick the boys and $5!$ ways to pick the girls, therefore there are $5!^2$ of the first type and $5!^2$ of the second type, so the final answer is $\frac{2(5!)^2}{10}=4!\cdot5!$

Answer 3

I have a different answer. there are no boys or girls that can seat together. Thus the first boy has 10 seats to take. The second boy already has 4 places, the 3d - 3, 4th -2 , 5th -1. So the boys can be distributed in $10*4*3*2=240$ ways. The girls can be distributed in $5!=120 ways$. So together it will be $240*120=28800$ ways to sit. Of course, if we don't count equivalent sitting up to rotation we should divide the value by 10. But still it is twice bigger than in the previous answer.