Probability: People sitting in a row (linear arrangement)

Question: Ten persons are seated at random in a row. What is the probability that a particular couple will be seated together? My attempt: 9! 2!/ 10! = $\dfrac{1}{5}$ , since there are 9! ways of sitting in pairs and 2! ways to arrange a couple.

The solution I'm given is $\dfrac{1}{63}$.

Can someone point out what I'm doing wrong?

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I get the same answer as you on first glance. Possibly a mistake? – Alex Becker Feb 28 '12 at 8:42
My reasoning: What you want to do is count the number of arrangements that have the couple seated together, and divide this by the total number of arrangements. In this case, we can treat seating $10$ people on a bench with the couple seated together as simply sitting $9$ people on the bench, putting the second person in the couple to the right of the first person in the couple, then possibly rearranging the two people in the couple. Hence $9!2!$ ways. Since there are $10!$ ways to seat $10$ people, the answer is $1/5$. – Alex Becker Feb 28 '12 at 8:43

You’re doing nothing wrong, assuming that your reasoning is similar to Alex Becker’s in his comment: the correct answer is indeed $\frac15$. Here’s another route to it.

There are $\binom{10}2=45$ pairs of seats, and the couple is equally likely to occupy any one of those $45$ pairs of seats. Nine of the $45$ pairs are adjacent, so the probability that they will occupy adjacent seats is $\frac9{45}=\frac15$, as you say.

And here is yet another. The man sits in an end seat with probability $\frac2{10}=\frac15$. If he’s in an end seat, only one of the remaining nine seats is adjacent to him, and his wife’s probability of getting that seat is $\frac19$. With probability $\frac45$ the man sits in one of the eight seats that have two neighbors, and in that case his wife’s probability of ending up next to him is $\frac29$. The overall probability that the end up sitting together is therefore

$$\frac15\cdot\frac19+\frac45\cdot\frac29=\frac9{45}=\frac15\;.$$

Added: And just for fun, here’s yet another. Imagine that the seats are arranged in a circle around a table. Wherever the wife is sitting, the husband’s probability of sitting next to her is $\frac29$. Then the table is taken away and the seats unwrapped into a straight line, with the breakpoint chosen at random: with probability $\frac1{10}$ it will fall between the husband and the wife, so with probability $\frac9{10}$ they will still end up together. Thus, they end up together with probability $$\frac9{10}\cdot\frac29=\frac15\;.$$

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And I just realized my second argument was the same as yours. – Alex Becker Feb 28 '12 at 8:58
@Alex: No harm done. Besides, I just added yet another sanity check! I’ll be damned if I can figure out, though, what chain of errors produced $1/63$: that factor of $7$ is downright mysterious. – Brian M. Scott Feb 28 '12 at 9:08
@Brain That's what gave me the courage to answer. The answer is clearly given by some factorial fraction, so how do you get $2\cdot 3\cdot 7$ in your answer? My guess is it was not a math error in the solutions, but rather the solution for a different problem was given. – Alex Becker Feb 28 '12 at 9:11
$1/945 = 1/(63 \times 15)$ is the probability that five couples find themselves sitting in their pairs. – Henry Feb 28 '12 at 10:14
${10 \choose 5} = 63 \times 4$. So $1/(63 \times 2)$ is the probability that five men and five women sit in single sex groups. – Henry Feb 28 '12 at 10:15

Your answer is correct. What you want to do is count the number of arrangements that have the couple seated together, and divide this by the total number of arrangements. In this case, we can treat seating 10 people on a bench with the couple seated together as simply sitting 9 people on the bench, putting the second person in the couple to the right of the first person in the couple, then possibly rearranging the two people in the couple. Hence 9!2! ways. Since there are 10! ways to seat 10 people, the answer is 1/5.

As a sanity check, here is another derivation. We concern ourselves only with the two people in the couple. There is a $8/10$ chance that the first person will not be on either end, and given this a $2/9$ chance that the second person will be next to them. There is a $2/10$ chance that the first person will be on the end, and given this a $1/9$ chance the second person will be next to them. Since these options are mutually exclusive, the overall probability is $$\frac{8}{10}\frac{2}{9}+\frac{2}{10}\frac{1}{9}=\frac{18}{90}=\frac{1}{5}$$

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That's great. Thanks a lot – eddflrs Feb 28 '12 at 9:23

You're right.

The number of pairs you can make is indeed $10 !$, and to get the number of combinations in which you can arrange the desired pair is as follows:

Let's have ${a,b,c,d,e,f,g,h}$ the 8 unimportant people and ${i,j}= {Icarus", Jessica"}$ the desired pair.

Now we can see the problem as a word-formation problem. We want a 10 characters word containing $ij"$ together. Later we multiply by two to consider the case $ji"$. So the trick is to notice FIRST how many 8-letter words we can make $= 8!$

We procede to insert the $ij"$ into the formed word: notice the 9 spaces between each letter where we can insert it (remember to count beginning and end). This is $\binom {9} {1}$

Putting it all together, we have $P[`ij" together]=\frac{2* 9* 8!}{10!}$ which is equal to $1/5$

Probably a type-o or a mistake from the book/homework.

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Expanding Brian Scott's idea for the general case, you can end up with a very handy and easy formula, given a row of $n$ chairs:

$$\text{Number of pairs of seats} = {n \choose{2}} = \frac{n!}{(n-2)! 2!} = \frac{n(n-1)}{2}$$ $$\text{Number of contiguous pairs of seats} = n-1$$ Therefore, $$P(\text{Choosing contiguous pairs among n seats}) = \frac{n-1}{\frac{n(n-1)}{2}} = \frac{2}{n}$$ So, in your specific case, $$P(\text{Choosing contiguous pairs among 10 seats}) = \frac{2}{10} = \frac{1}{5}$$

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PS: Trying to use this formula as a conversation starter with the cute person sitting next to you in a row of n chairs will likely cause said person to stand up and search a non-contiguous option. – kaiser_soze Aug 17 '15 at 0:27

There are $10!$ ways how to put these $10$ persons in the row, the couple can seat this way $12$, $23$, $34$, $45$, $56$, $67$, $78$, $89$, $10 \ 1$ or $21$, $32$, $43$, $54$, $65$, $76$, $87$, $98$, $1\ 10$ so $20$ possibilities but there are $8$ other persons, there are $8!$ possibilities to seat them so the probability is

$$P = 20 \cdot \frac{8!}{10!} = \frac{2}{9} = 22.2\%$$

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10-1 and 1-10 do not seem to fit the problem statement. And your answer does not match the "given" one either. – ronno Oct 27 '12 at 18:51
You forgot to mention $9 \ 10$ and $10 \ 9$ as seatings, but as ronno said, the $10 \ 1$ and $1 \ 10$ don't fit the problem statement (the people are sitting in a row, not in a circle). So you only get $18$ possibilities, also giving $2/10 = 20\%$. And welcome to math.SE! – TMM Oct 27 '12 at 19:38

protected by Zev ChonolesDec 9 '15 at 10:05

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