# Probability question about married couples

If four married couples are arranged in a row, what is the probability that no husband sits next to his wife?

Would it be

$1- \frac{2(4!)}{8!}$?

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How do you get $2(4!)$? – Arturo Magidin Sep 29 '11 at 16:06
There are 2 ways the husband and wife can arrange themselves and 4! ways to order the the four husband/wife pairs, so there are 2*4! ways where the husband is next to the wife. – lord12 Sep 29 '11 at 16:08
If all four couples are sitting next to each other, then you have $4!$ ways of arranging the couples, and $2$ ways of arranging each couple; that would be $2^4(4!)$ ways. But "all four couples are sitting next to each other" is not the complement of "no husband is sitting next to his wife"; you could have just one couple sitting together and the other three not. – Arturo Magidin Sep 29 '11 at 16:09
Do you require that men and women alternate? That will impact the probability. Also, if you seat the couples as couples, the probability is zero. – Ross Millikan Sep 29 '11 at 16:36

Let us first the label the couples as $A1,A2,A3$ and $A4$.Define the events $E1,E2,E3,E4$ as:

E1 = Event of A1 sitting together.
E2 = Event of A2 sitting together.
E3 = Event of A3 sitting together.
E4 = Event of A4 sitting together.


Hence,the probability $P\space(E1 \cup E2 \cup E3 \cup E4)$ will give the probability of the case where at least one of the four couples will sit together.

So, your required answer is $1-\space$$P\space(E1 \cup E2 \cup E3 \cup E4) PS:We can use the principle of mutual inclusion and exclusion to find P\space(E1 \cup E2 \cup E3 \cup E4) and it should give \frac{23}{35} (if I haven't made any mistake with calculation). - oops: sorry and never mind, I'm getting the same answer. I was comparing it to the answer to the actual question, not the computation of the latter probability. I think you're right, then. – Arturo Magidin Sep 29 '11 at 17:04 @Arturo Magidin:Thanks,so according to my approach 1-\frac{23}{35} =\frac{12}{35} should be the correct answer. – Quixotic Sep 29 '11 at 17:10 Assuming you seat the 8 individuals at random, one way of doing this is to use the inclusion exclusion principle and turn a number of couples into virtual individuals so they must sit next to each other to get$$1-\frac{ 2^1{4 \choose 1} 7! - 2^2{4 \choose 2} 6! + 2^3{4 \choose 3} 5! - 2^4{4 \choose 4} 4!}{8!}$$- Upvote for simplest solution. What would be your simplest solution for seating in a circle ? – true blue anil May 31 '15 at 19:01 @trueblueanil: Since there are (modulo rotation) 7! ways of arranging 8 people around a table, I would say 1-\dfrac{ 2^1{4 \choose 1} 6! - 2^2{4 \choose 2} 5! + 2^3{4 \choose 3} 4! - 2^4{4 \choose 4} 3!}{7!}. I have not checked this but for 2 couples the equivalent statement seems to give \frac13 which seems to be correct. – Henry May 31 '15 at 21:56 This is great. So much simpler than Bogart and Doyle's solution to the relaxed menage problem. math.dartmouth.edu/~doyle/docs/menage/menage/menage.html May be a simpler solution to the menage problem also can be evolved starting from here. – true blue anil Jun 1 '15 at 4:23 @trueblueanil: I think my answer is pretty much the same as Bogart and Doyle, where my "turn a number of couples into virtual individuals" corresponds to their "non-overlapping dominos" – Henry Jun 1 '15 at 7:07 Of course, the paper is a classic, but by treating the circle as unnumbered, and not introducing an intermediate d_k, it becomes simpler for the relaxed menage problem. – true blue anil Jun 2 '15 at 6:08 Let's count the complement. If at least one couple stands together, you have \binom{4}{1} = 4 ways of choosing the couple, 2 ways of arranging them, and 7 positions in which to place them (the first person in the couple cannot stand in the last place of the line), and the remaining 6 positions can be filled in 6! ways; this gives a total of 4\times 2\times 7\times 6! = 8!=40320 ways. However, this overcounts by counting arrangements in which more than one couple sits together twice. If at least two couples stand together, we have 4\times 3 ways of choosing the first and second couple, 4 ways of arranging them, 15 ways to stand them in the line in that order, and 4! of filling in the remaining 4 places, for a total of 4\times 3\times 4\times 15\times 4! = 17280 ways. If at least 3 couples stand together, we have 4\times 3\times 2 ways of choosing the couples (in order), 8 ways of arranging them amongst themselves, 10 ways of arranging them in the line, and 2 ways of filling the remaining seats, for a total of 4\times 3\times 2\times 8\times 10\times 2= 3840 ways. If all four couples sit together, we have 4\times 3\times 2\times 1 way of arranging the couples, and 2^4 = 16 ways of arranging each couple, for a total of 384 ways. So, let's see: when we count "at least one couple stands together", we count the arrangements with exactly one couple together once, the arrangements with exactly two couples together twice; the arrangements with exactly three couples three times, and the arrangements with all four couples together four times. When we could "at least two couples together", we count the arrangements with exactly two couples together once, the arrangements with exactly three couples together \binom{3}{2}=3 times; and the arrangements with all four couples together \binom{4}{2}=6 times. When we could "at least three couples together", we count the arrangements with exactly three couples together once, and the arrangements with all four couples together \binom{4}{3}=4 times. So if we take:$$(\text{at least one}) - (\text{at least two}) + (\text{at least three}) - (\text{all four})$\$ we get the count exactly right.

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