Posted by Sarah on Tuesday, October 2, 2012 at 9:20pm.
1. Romeo, Juliet, _ _ _ _
2. Juliet, Romeo, _ _ _ _
3. _ Romeo, Juliet _ _ _
4. _ Juliet, Romeo _ _ _
5. _ _ Romeo, Juliet _ _
6. _ _ Juliet, Romeo _ _
7. _ _ _ Romeo, Juliet _
8. _ _ _ Juliet, Romeo _
9. _ _ _ _ Romeo, Juliet
10._ _ _ _ Juliet, Romeo
There are 5*2 possibilities for Romeo and Juliet.
If we have a remaining seats with 4 in a row, (numbers 1, 2, 9 and 10 for Romeo and Juliet=4 possibilites for Romeo and Juliet), we can fill the rest
Caesar _ Brutus _
Brutus _ Caesar _
Caesar _ _ Brutus
Brutus _ _ Caesar
Each of these has two possibilities for homer and Pierre, so there are 4*4*2 = 32 possibilities for everyone when Brutus, homer, Pierre and Caesar sit 4 in a row
If three of Homer, Pierre, Brutus, and Caesar sit next to each other (numbers
3, 4, 7, 8 for Romeo and Juliet = 4 possibilities for Romeo and Juliet), we have
Caesar, Rom, Juliet, Brutus _ _
Caesar, Rom, Juliet, _ Brutus _
Caesar, Rom, Juliet, _ _ Brutus
Brutus, Rom, Juliet, Caesar _ _
Brutus, Rom, Juliet, _ Caesar _
Brutus, Rom, Juliet, _ _ Caesar
_ Rom, Juliet, Caesar _ Brutus
_ Rom, Juliet, Brutus _ Caesar
For each of these, there are 2 possibilites for Homer and Pierre
8*4*2 = 48 for these arrangements
For arrangements 5 and 6 for Romeo and Juliet, we have
Caesar _ Rom, Juliet, _ Brutus
Caesar _ Rom, Juliet, Brutus _
Brutus _ Rom, Juliet, Caesar
Brutus _ Rom, Juliet, _ Caesar
4*2*2 = 16 possibilities
Total # of possibilities is 16 + 32 + 48 = 96
I hope I did this right!!
2-39=