# In how many way can $7$ girls be seated at a round table so that $2$ particular girls are separated?

## Question

In how many way can $7$ girls be seated at a round table so that $2$ particular girls are separated?

## My Approach

$6!$ ways for the girls to sit at a round table.

which $2$ girls will be separated $?\Rightarrow \binom{7}{2}$

Let these $2$ girls sit together $5! \cdot \binom{7}{2}$

Required answer=$6!-5! \cdot \binom{7}{2}$

Am I right?

• I'd expect that they mean two specified girls. That is, you don't need to choose $2$ from the $7$, the two girls are already identified. Note: you need to consider the blocks $AB$ and $BA$. – lulu Oct 3 '17 at 19:32
• so it should be $6!-5!$.? – laura Oct 3 '17 at 19:33
• Well, you are neglecting my comment about $AB,BA$. You have to consider both orders for the two girls. – lulu Oct 3 '17 at 19:36
• okk so i should take $5!$ and permute $AB,BA$ among themselves i.e $2!$ i.e $5!*2!$? – laura Oct 3 '17 at 19:41
• Yes, that's right. So, $6!-2\times 5!$ – lulu Oct 3 '17 at 19:42

You do not get to choose the two girls who are separated, so there should not be a $\binom{7}2$ term. There are two girls named Mary and Paula, say, and you want to compute the number of ways to seat all the girls so Mary and Paula are not next to each other.

It is a good idea to subtract the cases where Mary is next to Paula. A good way to count the cases where Mary is next to Paula is this: imagine that Mary and Paula are glued together into a single person. There are now six people instead of seven, who can be put in a circle in $5!$ ways. There are also two choices for how to orient the Mary/Paula pair, for a total of $5!\cdot 2$ ways.

$6! - 5!{7\choose 2} < 0$ hence not correct.

I would say seat one of the troublesome girls first. It doesn't matter where, it is a round table.

There are $4$ places to seat the friend / enemy.

and $5!$ to seat the rest.

$4\cdot 5! = 480$