1
$\begingroup$

If three married couples (so 6 people) sit in a row of six seats at random, what is the probability that no couples sit together?

Another way to think about it (couples are AB, CD, and EF)

enter image description here

$\endgroup$
  • $\begingroup$ That looks like a small enough problem to just try and draw a few possibilities ... Did you try that? $\endgroup$ – Bram28 Oct 31 '17 at 21:01
  • $\begingroup$ What have you attempted? Where are you stuck? $\endgroup$ – N. F. Taussig Oct 31 '17 at 21:36
  • $\begingroup$ I added the work that I had done... this is similar to Bram's solution except with distinct values for each couple. I'm wondering if there is a way to think about it without actually making a tree of possible values... $\endgroup$ – Murey Tasroc Nov 1 '17 at 2:46
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Nov 1 '17 at 12:33
2
$\begingroup$

Just see how you can have no couples sitting together.

So let's say we have $112233$ that we need to mix up so that no two of the same are next to each other.

Start with two different ones:

$$12$$

The next can be a $1$, and then the rest is forced: $121323$

The next can also be a $3$, and then you have a few options: $123123, 123132, 123213, 123231$

OK, so that's $5$ options, but since the first two can be $12, 21, 13, 31, 23,$ or $32$ you get $6 \cdot5 = 30$ ways for this to happen .... out of $\frac{6!}{2!2!2!}=90$ total ways... giving you a probability of $$\frac{30}{90}=\frac{1}{3}$$

$\endgroup$
  • $\begingroup$ Why would you consider a couple to be the same. Ie, why not thinking of a couple as HW. $\endgroup$ – Sul Oct 31 '17 at 21:44
  • $\begingroup$ @SxS Because this way it simplifies the math a bit. $\endgroup$ – Bram28 Oct 31 '17 at 21:45
  • $\begingroup$ Ah I see, but generally, how could one guarantee that this simplification would not affect the final answer? $\endgroup$ – Sul Oct 31 '17 at 21:49
  • $\begingroup$ @SxS Distinguishing between the members of each couple would multiply both the numerator and denominator by $8$, which has no effect on the probability. $\endgroup$ – N. F. Taussig Oct 31 '17 at 21:50
  • $\begingroup$ @N.F.Taussig Ah right, thanks guys! $\endgroup$ – Sul Oct 31 '17 at 21:51
1
$\begingroup$

There are $6!$ possible seating arrangements. From these, we must exclude those in which one or more couples sit in adjacent seats.

There are three ways to select a couple who sit in adjacent seats. That gives us five objects to arrange, the couple and the other four people. The objects can be arranged in $5!$ ways. The couple that sits together can be arranged internally in $2!$ ways. Hence, there are $$\binom{3}{1}5!2!$$ seating arrangements in which a couple sits in adjacent seats.

However, if we subtract these seating arrangements from the total, we will have subtracted too much since we have counted seating arrangements in which two couples sit together twice, once for each way we could designate one of the couples as the couple that sits in adjacent seats. Since we only want to subtract such couples once, we must add them back.

There are $\binom{3}{2}$ ways to select two couples that sit together. That gives us four objects to arrange, the two couples and the two other people. The objects can be arranged in $4!$ ways. Each of the two couples that sit in adjacent seats can be arranged internally in $2!$ ways. Hence, there are $$\binom{3}{2}4!2!2!$$ seating arrangements in which two couples sit together.

When we subtracted arrangements in which a couple sits together, we counted seating arrangements in which all three couples sit together three times, once for each way we could have designated one of those couples as the couple that sits together. When we added arrangements in which two couples sit together, we counted seating arrangements in which all three couples sit together three times, once for each of the $\binom{3}{2}$ ways we could have designated two of the three couples as the ones that sit together. Therefore, we have not excluded seating arrangements in which all three couples sit together at all.

There are $3!$ ways to arrange three couples. Each couple can be arranged internally in $2!$ ways. Hence, the number of seating arrangements in which all three couples sit together is $$\binom{3}{3}3!2!2!2!$$

By the Inclusion-Exclusion Principle, the number of seating arrangements of the three couples in which no couples sit together is $$6! - \binom{3}{1}5!2! + \binom{3}{2}4!2!2! - \binom{3}{3}3!2!2!2!$$

The probability that no couple sits together is $$\frac{6! - \dbinom{3}{1}5!2! + \dbinom{3}{2}4!2!2! - \dbinom{3}{3}3!2!2!2!}{6!} = 1 - \frac{\dbinom{3}{1}5!2! - \dbinom{3}{2}4!2!2! + \dbinom{3}{3}3!2!2!2!}{6!}$$

$\endgroup$
0
$\begingroup$

There are $6!$ ways of seating $6$ people ${H_1, W_1, H_2, W_2, H_3, W_3}$ in a row of six seats.

Let's now treat each couple as a unit $U$, so we've got $$U_1 = H_1, W_1$$ $$U_2 = H_2, W_2$$ $$U_3 = H_3, W_3$$

Now, there are $2*4!$ ways of seating $3$ units of two people in row of $3$ seats.

It follows that there are $6! - 2*4! = 672$ ways in which no couple sit nex to each other.

Hence the probability is $$\frac{672}{6!} = \frac{14}{15}$$

This is how I, personally, would go about it.

$\endgroup$
  • 1
    $\begingroup$ Your answer is incorrect. All couples sitting together is not the opposite of no couples sitting together. What you need to eliminate is the possibility that at least one couple sits together. $\endgroup$ – N. F. Taussig Oct 31 '17 at 21:35
  • $\begingroup$ @N.F.Taussig oh yeah. My bad. $\endgroup$ – Sul Oct 31 '17 at 21:41
  • 1
    $\begingroup$ I have written an answer that demonstrates how to exclude the possibility that at least one couple sits together. $\endgroup$ – N. F. Taussig Nov 1 '17 at 12:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.