Different seating arrangements on a 6 seater

Question

Six actors sit in a row to have their photographs taken. Romeo and Juliet insist on sitting next to each other. Caesar refuses to sit next to Brutus. Falsta and Puck don't mind where they sit. How many completely different ways altogether, can they be seated?

Answer 1

This is kind of a brute force approach, but it works pretty well since there aren't too many options.

Romeo and Juliet are represented by Q, Brutus and Caesar are represented by X, and Falsta and Puck are represented by ?. We will now list all the possible seating arrangements. Note that each of the pairs are interchangeable, so we will multiply our final result by 8.

QQX?X? | QQX??X | QQ?X?X

XQQX?? | XQQ?X? | XQQ??X | ?QQX?X

X?QQX? | X?QQ?X | ?XQQX? | ?XQQ?X

I have listed 11 arrangements here. The first 7 arrangements have their mirror counterparts where the QQ are on the right side of the 6 seats, so there are really 11+7=18 arrangements I should have listed. Multiplying this by 8 gives a total of 144 possibilities.

It's entirely possible I have neglected something, so someone should check my work.

EDIT: I think a good check is that I ended up with 18 arrangements, which only has one factor of 2. (i.e., I can only divide 18 by 2 wholly once.) This means that I have accounted for all of the symmetries -- with the one I included in the 18 being the reflective symmetry of the seating arrangements. (i.e., if one seating arrangement works, then the complete reverse order works too).

Answer 2

Romeo and Juliet have internally two ways to sit. If we take that into account we can for all practical purposes glue them together treat them as one. We are then down to 5 "people". So now we consider Caesar, Brutus, Falsta, Puck and Julieo. Caesar and Brutus sitting together is internally two ways and externally (just four people-hybrids now) $2 \cdot 4!$. We need to remove that from the $5!$ if all 4 and R-J hybrid were allowed to sit in any order. So all in all:

$$2\cdot(5! - 2\cdot4!) = 144$$

Note this includes counting if Romeo is on the left and Julia on the right or vice versa.

Answer 3

Another way:

Romeo and Juliet can be glued together as $RJ$ or $JR:\;\; 2$ ways

Leave out Brutus and Caesar for the moment, then there are $3$ entities, permutable in 3! ways

$\Large\uparrow\bullet\Large\uparrow\bullet\Large\uparrow\bullet\Large\uparrow$ with $4$ gaps where Brutus and Caesar can be inserted in $4\times$3 ways.

Putting everything together yields $2\times3!\times4\times3 = 144$