If 6 married couples and 3 unmarried persons are arranged in a row, find the probability that no husband sits next to his wife?


closed as off-topic by Michael Hoppe, Claude Leibovici, user63181, Shuchang, John Habert Mar 27 '14 at 13:58

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  • $\begingroup$ gay marriage is on advance, so the task should be more precise :=) $\endgroup$ – mathse Mar 27 '14 at 12:39

To find a probability, we need to divide the number of ways where at least 1 couple is together by the total number of seating arrangement. This probability will be equal to the chance that at least 1 couple is together; from there, we can easily find the probability for the case were no couple are together.

We know we have a total of 15 people present in the row. To make the question simpler, let us assume each person has an identity. So overall, the way to mix our people up is really simple:


(15 possibilities for first seat, times 14 possibilities for second seat, times 13 possibilities for third seat, etc.)

Now when we have a couple together, essentially, we can regard them as one unit, and as long as they are "touching", they can be permuted as desired. Hence for the scenario when an arbitrary couple is together, we have 14! permutations.

Now note: The 14! permutation regards the positioning of the couple themselves (on their 2 seats) as arbitrary, i.e. non ordered. However, on the 15! permutations, a swap represents an alternative permutation. So to bring the two permutation counts to the same "world" so that we can extract the quotient, we need to multiply 14! by 2.

Notice how I said an "arbitrary" couple. Well, fortunately, we can count the number of couples that exist. Namely 6 (as there are 6 husbands and 6 wives). So we multiply 14!*2 by 6.

Wait though, we are not done! Consider: for the second couple's 14! permutations, some of them will have the previous couple together; this will be problematic, as those permutation will have already been counted by the previous 14! (the previous term will have accounted for that permutation). We can resolve this issue for the second couple by subtracting 13!. Why? Because with 13!, we are essentially grouping the 2 couples into 2 units, and the 13! will give the number of permutations for which this will happen.

Similarly for the 14! of the third couple, some permutations will have the 1st and 2st couple "touching", these will be permutations that have already been counted, so we subtract 2*13!. Now note, we will be subtracting too much, as for some of these subtractions, all three couples will be "together" and we will therefore be subtracting these permutations twice, so we add 12! (we have grouped 3 couples as one (15-3 = 12) so that the permutations are "saved" from annihilation).

For the fourth 14!, we will have -3*13! + 2*12! - 11!.

For the fifth (couple's) 14!, we will have -4*13! + 3*12! - 2*11! + 10!

And of course, for the sixth 14!, we will have -5*13! + 4*12! - 3*11! + 2*10! - 9!

Now, all we have to do is to take the quotient (with the above corrections included):

2(6*14! - 15*13! + 10*12! - 6*11! + 3*10! - 1*9!)/15!

This is about 0.66 (0.66412..)

so 66% of the time, two couples will be next each other. By deduction 34% of the time, no couples will be together.


  • 1
    $\begingroup$ very nice :) {} {} $\endgroup$ – mookid Mar 27 '14 at 13:27

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