A small airplane has rows of seats with seats in each row. Eight passengers have boarded the plane and are distributed randomly among the seats. A married couple is next to board. What is the probability there will be 2 adjacent seats in the same row for the couple?
Suppose the passengers are indistinguishable. There are 
total ways to distribute the passengers. We proceed with complementary counting, and instead, will count the number of passenger arrangements such that the couple cannot sit anywhere. Consider the partitions of among the rows of seats, to make our lives easier, assume they are non-increasing. We have 
.
For the first partition, clearly the couple will always be able to sit in the row with $" width="8" height="12" />$" width="8" height="12" />$" width="8" height="12" /> occupied seats, so we have $" width="8" height="12" />$" width="8" height="12" />$" width="8" height="12" /> cases here.
ways to permute the partition. Now the rows with exactly passenger must be in the middle, so this case generates cases.
For the third partition, there are 
ways to permute the partition. For rows with passengers, there are 
ways to arrange them in the row so that the couple cannot sit there. The row with passenger must be in the middle. We obtain cases.
For the fourth partition, there is way to permute the partition. As said before, rows with passengers can be arranged in ways, so we obtain cases.
Collectively, we obtain a grand total of 
cases. The final probability is 
= \boxed~\frac>$" width="156" height="49" />.
Suppose the passengers are indistinguishable.
What this question is asking, is really, if 4 empty seats are placed, what is the probability that there are 2 adjacent seats open.
We proceed by casework.
cases per pair of open seats for total cases.
Case 2: There is one row of open seats. ways to choose the row and to choose the final empty seat for cases.
cases.
In total, we get 
cases total for a probability of ![\[\frac>=\frac=\boxed<\mathbf<(C)>~\frac>\]](https://latex.artofproblemsolving.com/5/f/a/5fa7168f21221853eddf5f28a64cc94595cec522.png)
We notice that if we have a middle seat in a row, then the couple cannot sit in that row. So, we perform complementary casework.
Case 1: Four people sitting in middle seats.
total combinations for this case.
Case 2: Three people sitting in middle seats.
In this case, there are 
ways to permute the rows in which the middle seat is occupied. For the row in which the people do not occupy the middle row, we must have two people sitting at the ends of the rows to guarantee the couple cannot sit there. So, for the rest of the 3 people, there are 6 possible seats. So, there are 
total combinations.
Case 3: Two people sitting in middle seats.
In this case, there are 
ways to permute the rows in which the middle seat is occupied. For the rows in which the people do not occupy the middle row, we must have two people sitting at the ends of the rows to guarantee the couple cannot sit there. So, for the rest of the 2 people, there are 4 possible seats. So, there are 
total combinations.
Case 4: One person sitting in a middle seat
In this case, there are 
ways to permute the rows in which the middle seat is occupied. For the rows in which the people do not occupy the middle row, we must have two people sitting at the ends of the rows to guarantee the couple cannot sit there. So, for the rest of the last person, there are 2 possible seats. So, there are 
total combinations.
Case 5: Zero people sitting in a middle seat
In this case, we must have every person sitting at the ends of the seats. So, there is only 1 combination.
In total, we have
![\[\dbinom<8></p>
<p> + \dbinom \cdot \dbinom + \dbinom \cdot \dbinom + \dbinom \cdot \dbinom +1\]](https://latex.artofproblemsolving.com/5/3/1/531b8d58558672256d0fa07e1f2baa44101924de.png)
. So, finally employing complementary counting, we have that the probability that there will be 2 adjacent seats for the couple is
![\[1 - \dfrac<195></p>
<p> = \dfrac.\]](https://latex.artofproblemsolving.com/c/e/6/ce6ec7638d7255c99580a9ca73ca164f2aa02b9e.png)
There are for two people (the married couple) to be seated. This will be our denominator.
There are pairs of seats that are next to each other in the diagram ( per row; left-middle and middle-right). This will be our numerator.
Since there are total people on the plane, we should multiply our numerator by that to account for all ways the 10 people could be seated (e.x. the husband and the wife could be switched around and it would still work, same applies to the other passengers)
Therefore, our numerator is .
This creates the fraction 
, which simplifies to ![\[\boxed<(\text<\bf<C>>) \: \frac>.\]](https://latex.artofproblemsolving.com/6/8/b/68b00f170a12d49a010d2c069dacbe0788ada616.png)
- Siddharth Mirchandani (svm2020), John Adams Middle School
Remark: This solution is flawed because if there are people on the plane, the probablity could not be calculated this way, because intuitively, the probability should be decreased (as the probability for 9 people and 1 couple is 
), not increased (to be 
). -ericz
Consider the couple seated together and there should be 8 seated ways (2 ways in each row). And the other 8 people can be seated in other 10 seats randomly.
There will be total
Consider two double counting cases
Case I: the other 8 people are seated in (1, 3) (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3)
It was double counted for couple's seats (1, 1) (1, 2) and (2, 1), (2, 2)
Case II: one whole row is empty and the other 8 people are randomly seated in other rows
It was double counted for couple's seats such as (1, 1) (1, 2) and (1, 2) (1, 3)
There will be
- Orlando Liu Cupertino Middle School
