# round table 2

• Feb 18th 2013, 10:11 PM
kastamonu
round table 2
4 married couples sit at a round table. How many ways can they sit if each couple must sit seperately?

There is a formula for this solution but I couldn't get it.

Always, always, start a new thread for a new question.

Using inclusion/exclusion http://latex.codecogs.com/png.latex?...right%29%21%7D
• Feb 20th 2013, 06:23 AM
kastamonu
Re: round table 2
4 married couples sit at a round table. How many ways can they sit if each couple must sit seperately?

There is a formula for this solution but I couldn't get it. I am sorry I wanted to make some edits to my former post but I couldn't I just wanted to past the formula but I also pasted some messages with it I am sorry(Crying)
• Feb 24th 2013, 03:07 AM
russ123
Re: round table 2
I don't have a clue about the equation to solve this but I managed to do it brute force with Python.

Code:

```>>> import itertools >>> couples = {0:1, 2:3, 4:5, 6:7} >>> for k,v in couples.items():         couples[v] = k         >>> all_positions = list(itertools.permutations(range(8))) >>> def is_valid(position):         if position[0] == couples[position[-1]]:                 return False         for p in range(len(position) - 1):                 if position[p] == couples[position[p+1]]:                         return False         return True >>> all_positions = [p for p in all_positions if is_valid(p)] >>> len(all_positions) 11904 >>>```
seems to be 11904
• Feb 24th 2013, 04:07 AM
Plato
Re: round table 2
Quote:

Originally Posted by russ123
I don't have a clue about the equation to solve this but I managed to do it brute force with Python.
Code:

```>>> import itertools >>> couples = {0:1, 2:3, 4:5, 6:7} >>> for k,v in couples.items():         couples[v] = k >>> all_positions = list(itertools.permutations(range(8))) >>> def is_valid(position):         if position[0] == couples[position[-1]]:                 return False         for p in range(len(position) - 1):                 if position[p] == couples[position[p+1]]:                         return False         return True >>> all_positions = [p for p in all_positions if is_valid(p)] >>> len(all_positions) 11904>>>```
seems to be 11904

That is a gross over-count. There are only $7!=5040$ ways to seat eight people at a around table.
• Feb 24th 2013, 09:55 PM
russ123
Re: round table 2
Sorry, I didn't know the exact positions of the seats didn't matter, that only the relative positions mattered then I can see how it's 7!

--- edit ---

I tried to correct my mistake and got a much lower figure: 1488
...but I wonder if it's right this time. If so then this will tell you the answer for any number of couples

Code:

```from itertools import permutations def check(config):     '''     return True if no couples are next to each other     False otherwise     '''     if config[0] == 0 or config[-1] == 0:         return False     else:         for x in range(len(config) - 1):             if config[x] == config[x + 1]:                 return False     return True while True:     couples = int(raw_input("couples: "))     all_configs = [0] + (range(1, couples) * 2)     all_configs = [c for c in list(permutations(all_configs)) if check(c)]     print len(all_configs)```