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Posted by **Sarah** on Tuesday, October 2, 2012 at 9:20pm.

- Math -
**Jennifer**, Wednesday, October 3, 2012 at 11:41am1. 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!!

- Math -
**nayeli**, Tuesday, April 30, 2013 at 6:12pm2-39=