Math
posted by Sarah on .
Six people are going to sit in a row on a bench. Romeo wasnts to sit next to Juliet. Caesar does not want to sit next to Brutus. Homer and Pierre can sit anywhere. In how many ways can these people be seated?

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!! 
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