There are 3 married couples should sit in a straight row of 6 seats. If couples sit random, what is the chance that at least one man sits next to his wife?

And, what is the chance that exactly $3$ ($2$, and $1$ also) couples are thogether?


Total number of combinations: $6!=720$

I want to counting $3$ disjoints sets.

$A_1$: the number of combinations in which there are exactly 3 men next to his wifes, denoting each couple by $P_i$ we have $3!$ possible combinations $P_1 P_2 P_3, P_2P_3P_1,$ etc.. But each couple can order in $2!$ forms then:


$A_2$: the number of combinations in which there are exactly 2 men next to his wife, denoting each couple by $P_i$ we have $4!-3$ possible combinations if we fix $2$ couples $P_1, P_2$, since there are $2!3$ of $4!$ combinations in which the third couple is not togheter namely $P_1 a P_2 b$, $a P_1 b P_2$ and $a P_1 P_2 b$ and changing $P_1$ with $P_2$. And since each couple can ordered in $2!$ forms we need to add factors $2!2!2!$ like previous case. Then

$A_2=2!2!2!3(4!-2!3)=432$ But im not sure!!!

There is an error in $A_2=2!2!2!3(4!-2!3)=432$, since $2!2!3$ of $4!$ was favorable cases. Then: $A_2=2!2!3(2!2!3)=2!^43^2=144$ as noted Brian.

$A_3$: the number of combinations in which there are exactly 1 man next to his wife. I have not idea how to compute this monster.

$A_3=288$ solved by Brian. $A_1+A_2+A_3=480$


For $A_2$ you have a good idea, but you missed a detail. There are $3$ ways to choose which couple does not sit together; this is the factor that you missed. Then there are $3$ ways to split that couple (your $P_1aP_2b$, $aP_1bP_2$, and $aP_1P_2b$), $2$ ways to decide which couple is $P_1$ and which is $P_2$, and $2$ ways to seat the members of each couple, so


In terms of your calculation, that changes $4!-2!\cdot3$ to $4!-3\cdot2!\cdot3=6$, and $2!^3\cdot3\cdot6=144$.

Probably the easiest way to calculate $A_3$ is to subtract the total of the other three possibilities from $6!$, since you already have them.

Added: Since there are only three couples, $A_4$ isn’t too hard to calculate by hand. Let $A$ be the person in the first seat in the row; there are $6$ choices for $A$. Let $A'$ be $A$’s spouse. Anyone except $A'$ can sit in the second seat, so there are $4$ possibilities; call the person who sits there $B$. Now split the count into two cases.

  1. The third seat is taken by $C$, a member of the third couple. There are $2$ ways to choose $C$, so there are altogether $6\cdot4\cdot2$ ways to fill the first three seats with members of three distinct couples. The fourth seat can be occupied by $A'$ or $B\,'$, and in either case the last two people can sit in either order, so there are $2^2=4$ ways to fill out the row, for a total of $6\cdot4\cdot2\cdot4=192$ arrangements.

  2. The third seat is taken by $A'$, so that we have $ABA'$ in the first three seats. The rest of this calculation is spoiler-protected to give you a chance to finish it on your own.

$A'$’s spouse is in the first seat, so at first sight it might appear that any of the $3$ remaining people can take the fourth seat. However, if $B\,'$ takes it, the third couple will be forced to sit together, so it must actually be taken by a member of the third couple, say $C$, and the last three seats must be filled $CB\,'C'$: the only choice is in the order of the two members of the third couple. This case therefore accounts for another $6\cdot4\cdot2=48$ arrangements. Combining results, we see that $A_4=192+48=240$.

  • $\begingroup$ Am I right with $A_1$? $\endgroup$ – Gaston Burrull Mar 23 '13 at 1:42
  • $\begingroup$ @Gastón: Yes, you are. $\endgroup$ – Brian M. Scott Mar 23 '13 at 1:43
  • $\begingroup$ Im trying to compute $A_3$ computing $A_4$ but I think that $A_4$ (0 couples togheter) is another monster. $\endgroup$ – Gaston Burrull Mar 23 '13 at 19:24
  • $\begingroup$ @Gastón: I’ve added a calculation of $A_4$, leaving part of it spoiler-protected. I’m sure that there are better ways to do it, but this problem is small enough for brute force to be effective, so I didn’t look too hard for a better approach. $\endgroup$ – Brian M. Scott Mar 23 '13 at 20:32
  • $\begingroup$ I liked your idea, thanks! I finished without spoiler and I checked after. $\endgroup$ – Gaston Burrull Mar 23 '13 at 23:30

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