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? 
 A: 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 $$\frac9{10}\cdot\frac29=\frac15\;.$$
A: 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 $$\frac{8}{10}\frac{2}{9}+\frac{2}{10}\frac{1}{9}=\frac{18}{90}=\frac{1}{5}$$
A: 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.
A: 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,
$$
P(\text{Choosing contiguous pairs among 10 seats}) = \frac{2}{10} = \frac{1}{5}
$$
A: 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
$$P = 20 \cdot \frac{8!}{10!} = \frac{2}{9} = 22.2\%$$
