Ten chairs are evenly spaced around a round table and numbered clockwise from $1$ through $10$. Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or across from his/her spouse. How many seating arrangements are possible?

$\mathrm{(A)}\ 240\qquad\mathrm{(B)}\ 360\qquad\mathrm{(C)}\ 480\qquad\mathrm{(D)}\ 540\qquad\mathrm{(E)}\ 720$

Solution For the first man, there are $10$ possible seats. For each subsequent man, there are $4$, $3$, $2$, and $1$ possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is $10\cdot 4\cdot 3\cdot 2\cdot 1\cdot 2 = \boxed{480}$.

I am not satisfied with this solution. Is there another way to do this? (I just don't like this solution's..solution. There are many ways to proof the Pythagorean Theorem, and while you have no logical objections to others, surely there's some you like better or less than others? It's like that. I don't think I "think" like this solution. In any case, I just want to explore some other solutions that make better sense in my mind.)

  • $\begingroup$ Understood. I am asking for another solution to the problem is all though. I will edit it and try to express my intentions better. $\endgroup$ – mathflair Jan 2 '16 at 1:24
  • $\begingroup$ Even if this were not a request for a more satisfactory solution of some kind, I would hope that you would include more context, such as your own thoughts about solving it. After all, a multiple choice problem will often have some short cuts, and hearing your thoughts of one kind or another may better inform a Reader's responses toward solutions that you find more pleasing. $\endgroup$ – hardmath Jan 2 '16 at 1:43
  • $\begingroup$ Perhaps you could give an example of how you "think" to a similar problem so we have a better idea of what you are after. Personally that answer is exactly how I think about the problem. $\endgroup$ – Ian Miller Jan 2 '16 at 2:07
  • $\begingroup$ I just wanted to see if there were other ways to do this problem, so no one had to tailor a solution to my mind! I may have come across that point confusingly. ^^; $\endgroup$ – mathflair Jan 2 '16 at 5:45

This is not very different from your solution but maybe it is more clear: Let the men be $m_1, m_2, ..., m_5$ and their arrangement be the (ordered) vector $A=(a_1, ..., a_5)$ Since the seats are labelled, the arrangement $(1,3,5,7,9)$ is different than $(3,5,7,9,1)$ etc. There are 10 ways to pick $a_1$ for man $m_1$. For each arrangement vector $A$ there are $4!$ permutations for men $m_2$ to $m_5$ and all the men arrangements are now $10\times 4!$

Without losing generality let's consider one men's arrangement $(1,3,5,7,9)$ For this case let's name the women as $w_i$, where $i$ is the seat number where their spouse is seated. Women can go to the even seats. In seat 2 you cannot have $w_1$ and $w_3$ and not $w_7$ either (across her spouse in seat 7). That leaves two choices for seat 2: $w_5$ or $w_9$. If you pick $w_5$ there is only one possible arrangement for the rest of the even seats to give: $(w_5, w_7,w_9,w_1,w_3)$. Same as you pick $w_9$, there is only one arrangement. Working the same way you can confirm that there are only two possible arrangements for the women as your solution states and get the answer $2\times10\times4!$


Another way

Permissible positions of males in relation to their spouses is either $+3$ or $-3$, e.g. $1-4$ or $1-8$

The chart below with females at odd positions makes it obvious that if the $+3$ option is chosen for one couple, it needs to be so for all couples, (and so, too for $-3$)

$1 - \color{green}4-\color{red}{8}\quad\;\; 1-\color{red}4-\color{green}8$
$3 - \color{green}6-\color{red}{10}\quad 3-\color{red}6-\color{green}{10}$
$5 - \color{green}8-\color{red}{2}\quad\;\; 5-\color{red}8-\color{green}{2}$
$7 - \color{green}{10}-\color{red}4\quad 7-\color{red}{10}-\color{green}4$
$9 - \color{green}2-\color{red}6\quad\;\; 9-\color{red}2-\color{green}6$

Thus $10\cdot4!$ with the alpha female at odd positions, ditto at even positions to finally yield

ans = $2\cdot 10\cdot4!$


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