Couples sitting together

Two couples and a single person are seated at random in a row of five chairs. What is the probability that at least one person is not beside his/her partner?

I tried classifying a permutation into 3 groups

(no couples sit) + (1 couple) + (2 couples) = $5!$

when 1 couple sits together

= 1st couple + 2nd couple - 1st and 2nd couple

$= (4!)(2) + (4!)(2) - (3!)(2)(2)$

when 2 of the couples sit together,

$= (3!)(2)(2)$

when none of the couples sit together

= 120 - 1 couple - 2 couple = 24

then the answer would be $24+72 = 96/120$

.8, can someone tell me what i did wrong?

• Why not try using $P_{\text{ At least one person is not beside his/her partner}}=1-P_{\text{ Everyone sitting beside their partner}}$? – Mohammad Zuhair Khan Aug 1 '18 at 4:02
• the answer is 0.4 there is an answer key,also, using your logic, we still get 1- (3!)(2)(2)/120 – SuperMage1 Aug 1 '18 at 4:04
• The book answer of $0.4$ is wrong. If you like, I can write down all the $96$ combinations that you have listed above. – астон вілла олоф мэллбэрг Aug 1 '18 at 4:24
• @астонвіллаолофмэллбэрг Actually check the answer below. – Mohammad Zuhair Khan Aug 1 '18 at 4:31
• I suspect the book is answering the question "what is the probability that no person is next to his/her partner?", which gives $0.4$. – Ross Millikan Aug 1 '18 at 4:47

You can divide this into $3$ groups like Couple $1$ sit together, Couple $2$ sit together and both Couple $1$ and Couple $2$ sit together.

First let us say

Couple $1$ as $X_1X_2$

Couple $2$ as $Y_1Y_2$ and the random person as $R$

Now let us consider the case where Couple$1$ and Couple $2$ sit together. $$\{X_1,X_2\}\{Y_1,Y_2\}\{R\}$$Therefore the arrangement would be $3!\times2!\times2!=24$ ways

Now exactly one of the couples sit together.

Let Couple $1$ sit together, then we have$$\{X_1,X_2\}\{Y_1\}\{R\}\{Y_2\}$$Therefore, the arrangement would be $4!\times 2!=48$ ways but these $48$ arrangements also include arrangements when two couples sit together, so the number of ways for Couple $1$ is $48-24=24$

Similarly, Couple $2$ sit together, then we have $$\{Y_1,Y_2\}\{X-1\}\{R\}\{X_2\}$$Therefore, the arrangement would be $4!\times 2!=48$ ways but these $48$ arrangements also include arrangements when two couples sit together, so the number of ways for Couple $1$ is $48-24=24$

The number of arrangements when at least on of the couples sit together is $\implies24+24+24=72$ ways.

Total number of ways of arranging $5$ people is $5!=120$ ways.

Now, the probability that at least on of the couples sit together is $\dfrac{72}{120}=\dfrac35$

The probability that neither of the couples sits together in adjacent chairs is $\implies1-\dfrac35=\dfrac25=0.4$

Edit:

The above answer is correct if neither of the couples sits together in adjacent chairs.

As @Ross Millikan mentioned in the comment. The probability that at least one is not beside his her partner is $\implies 1-\dfrac{\mbox{Everyone sitting together as couple}}{120}\implies 1-\dfrac{24}{120}\implies1-\dfrac25=0.8$

• We were asked for the chance at least one person is not next to his/her mate, so it should be the number of ways to have no couples together $(48)$ plus the number of ways to have one couple together $(48)$, which supports $0.8$ – Ross Millikan Aug 1 '18 at 4:36

Your count of exactly one couple together is actually a count of at least one couple together. There are $2\cdot 4!$ for the first couple to be together, as you say. When you double that for the second to be together and only subtract once the cases where both couples sit together you are still counting the cases that both couples sit together once. If you want the number of ways that exactly one couple can sit together it should be $2\cdot (2\cdot 4!-3!\cdot 2 \cdot 2)=48$. There are also $48$ ways that neither couple sits together.

This still supports $0.8$ as the chance that somebody is not next to his/her partner, because you just subtract the $24$ ways both couples can sit together from the full $120$ ways.