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

Would it be

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

Quixotic
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lord12
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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
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    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

3 Answers3

13

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 $$1-\frac{ 2^1{4 \choose 1} 7! - 2^2{4 \choose 2} 6! + 2^3{4 \choose 3} 5! - 2^4{4 \choose 4} 4!}{8!}$$

Henry
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  • 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. https://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 01 '15 at 04:23
  • @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 01 '15 at 07:07
  • 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 02 '15 at 06:08
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    Why you have done 7! ,6! 5! & 4! – Koolman Oct 17 '16 at 16:59
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    @koolman: if you turn one couple into a virtual individual, you get $7$ individuals (real or virtual), who can be arranged $7!$ ways. Similarly with more couples turned into virtual individuals – Henry Oct 17 '16 at 17:06
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    @Henry but there will be 8 individuals (4 couples x 2) – Koolman Oct 17 '16 at 17:12
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    @koolman: they are not quite fully individual if they are forced to sit next to each other – Henry Oct 17 '16 at 17:16
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    @Henry so why only 7! – Koolman Oct 17 '16 at 17:18
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    We have 8 places for every individuals – Koolman Oct 17 '16 at 17:21
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    @koolman: If you are asking about the round table, then you lose a degree of freedom for the rotational symmetry – Henry Oct 17 '16 at 17:25
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    @Henry no I am not asking of round table . I am asking for arrangement in a row. – Koolman Oct 17 '16 at 17:32
  • For married couples in a row, is there a formula if a number of strangers are interspersed, eg how many seating arrangements would be there if apart from the four couples, four strangers were interspersed. – true blue anil Apr 29 '22 at 08:01
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    @trueblueanil Just bump up the factorials by the number of strangers, e.g. $8!\to 12!$ – Henry Apr 29 '22 at 10:47
10

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).

Quixotic
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  • 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
9

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.

Arturo Magidin
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  • Did not understand how did you get 15 and 10 in 2nd and 3rd calculation, can you please explain – Hardik Gupta Aug 14 '18 at 03:53
  • @Hardikgupta: Did you see the date? It's been almost seven years since I wrote that. – Arturo Magidin Aug 14 '18 at 04:28
  • I understand that, however I really liked your answer expcept that two values. it's ok, i'll try to decode it if I can (actually already tried hard and hence commented) – Hardik Gupta Aug 14 '18 at 07:02
  • @Hardikgupta: Sigh... do you remember what you were thinking 7 years ago? Take the two couples. if the first couple stands in positions 1 and 2, then the second couple can stand in positions 3-4, 4-5, 5-6, 6-7, or 7-8. If first couple stands in 2-3, then second couple could be in 4-5, 5-6, 6-7, or 7-8. If first couple in 3-4, then 2nd in 5-6, 6-7, or 7-8; if first in 4-5, then second in 6-7 or 7-8; if first in 5-6, then second in 7-8. That's a total of 15 ways to arrange the two couples. – Arturo Magidin Aug 14 '18 at 14:45
  • "If first couple in 3-4, then 2nd in 5-6, 6-7, or 7-8" and 1-2. – pawelK Oct 07 '19 at 20:45
  • @ArturoMagidin I don't understand how you counted this - "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 you counted "at least one couple stands together". – Ankit Seth Jul 22 '21 at 16:11
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    @AnkitSeth: You count the same couple twice if two couples sit together: once when they are the first couple, once when they are the second couple. You count it three times when three sit together (same reasoning), etc. – Arturo Magidin Jul 22 '21 at 19:14
  • @ArturoMagidin Ok. That I understood. But I don't understand, how does this reasoning applies when you counted the arrangements with exactly three couples together $\binom {3} {2}=3$ times; and arrangements with all four couples together $\binom {4} {2}=6$ times when "atleast 2 couples sit together"? – Ankit Seth Jul 23 '21 at 02:35
  • @AnkitSeth: Thank you for asking me to reconstruct an argument from almost ten years ago. As you no doubt know, I remember everything I ever wrote as if it were just yesterday. The same arrangement with exactly two couples is counted once when I count "at least two couples", and twice when I count "at least one couple", once for each of the two couples. The same arrangement with exactly three is counted three times when I count "at least one" (once for each of the three), $\binom{3}{2}$ when I count "at least two" (once for each choice of two), and once for "at least three". Etc. – Arturo Magidin Jul 23 '21 at 03:11
  • @AnkitSeth: This is basic inclusion-exclusion. – Arturo Magidin Jul 23 '21 at 03:18
  • @ArturoMagidin Sorry, I am not that good at combinatorics. – Ankit Seth Jul 23 '21 at 04:19