# Sit People on a Bench Such That One is Always on the Left of the Other

I'm new to Combinatorics and would like to understand the idea of the following combinatorics questions:

In How many ways can you sit 4 couples on a bench such that:

1. Lucy always sits to the left of Carol (Not necessarily adjacent).

I was trying to count all the options for sitting Lucy (first place, second place and so on) but I'm sure there's more elegant way of doing that. Just to be sure the result I got is: 30,960 ways.

1. Lucy would never sit next to its partner and Carol would never sit next to its partner.
• For the first question, of all possible orderings, how would you compare the count of the ones with Lucy to the left of Carol to the count of the ones with Lucy to the right of Carol? – quasi Dec 29 '17 at 12:05
• I assumed you used $\binom 87 = 40320$. – Mohammad Zuhair Khan Dec 29 '17 at 12:05
• I presume it should be $7! = 5040$. – Mohammad Zuhair Khan Dec 29 '17 at 12:06
• Or it could be $7+6+5+4+3+2+1 = 28$. – Mohammad Zuhair Khan Dec 29 '17 at 12:07
• I put Lucy on the first spot. Then there are 7! options. Then I put Lucy in the second spot so there are: 6 options for the first spot, Lucy, 6 options for the third spot, then 5, 4, 3, 2,1. And then I put Lucy in the third spot and so on, eventually summing up all the possibilities. – Moshe King Dec 29 '17 at 12:10

2.

There are $2\cdot7!$ arrangements s.t. Lucy and partner sit together.

There are $2\cdot7!$ arrangements s.t. Carol and partner sit together.

There are $4\cdot6!$ arrangements s.t. [Lucy and partner] and [Carol and partner] sit together.

That gives $2\cdot7!+2\cdot7!-4\cdot6!$ arrangements s.t. [Lucy and partner] or [Carol and partner] sit together.

There are $8!$ arrangements in total, so:$$8!-2\cdot7!-2\cdot7!+4\cdot6!$$arrangements s.t. [Lucy and partner] do not sit together and [Carol and partner] do not sit together.

$\underline{Another\; approach\;for\;2}$

• Keep Lucy and partner separate by placing them in the gaps of the $6$ others,
$\;-1-2-3-4-5-6-\;$ and permute the others, thus $\;7\cdot6\cdot6!$ ways.

• Now subtract arrangements with Carol and partner together, denoting them as a super $C$ internally permutable in $2$ ways, $\; -1-\mathscr C- 3 - 4 - 5 - \;,$ thus $\;6\cdot5\cdot2\cdot5!$

• The final ans is $\;42\cdot6! - 60\cdot5! = 23,040$