Probability of 2 students sitting next to each other I got this problem on my recent quiz that I couldn't for the life of me figure out how to solve.

There is a classroom with 5 rows and 7 seats per row. The class has 31 students, and exactly 2 of the students have the same birthday. If the students are assigned seats randomly, what is the probability that the students who share a birthday will sit next to each other in the same row?


I got $0.0645$ but the answer supposed to come out to be $0.05042$, according to my professor. Any help would be appreciated!
 A: We can solve this using the law of total probability. The first student of the pair sits at one of the seats at the end of the rows with probability $2/7$. In this case, there is only one valid sit exists for the second student, resulting in probability $1/(35-1)$. The first student sits at an inner seat with probability $5/7$. In this case there are 2 valid options for the second student, which has probability $2/(35-1)$. So in total, the probability is
$$
\frac 27\cdot\frac1{34}+\frac 57\cdot\frac2{34}=\frac{6}{119}=0.0504202.
$$
Note that the number of students (so long as it is $\ge2$ and $\le35$) does not matter.
A: Let $p_{i,j}$ the probability that $i$ is the seat number of the first student and $j$ the seat number of the second. It's clear that:
$$p_{i,j} = \begin{cases}\frac{1}{35 \cdot 34} & \text{if}~i \neq j \\ 0 & \text{else}\end{cases},$$
since the number of possible allocations of the two (distinct) student is $35 \cdot 34.$
Consider the first row of chairs. There are combinations exactly $12$ such that $i$ and $j$ are seated side-by-side (by enumeration, you get $(1,2), (2,1), (2,3), (3,2), (3,4),(4,3),(4,5),(5,4), (5,6), (6,5),(6,7),(7,6))$. Considering $5$ rows, you have $12 \cdot 5$ possibility over $35 \cdot 34$, that is:
$$\frac{12 \cdot 5}{35 \cdot 34}  \simeq 0.050420168067227$$
A: In each row of seven seats there are 6 pairs of adjacent seats; times 5 rows in total gives 30 possible pairs of adjacent seats. Since the relative position of the two co-birthday students doesn't matter, that is a total of 60 possible ways in which the two co-birthday students could be seated beside each together as required.
There are 35 * 34 total possible seating arrangements for the two co-birthday students.
Therefore the solution is 60 / (35 * 34) ~ 0.05042
A: Imagining $35$ seats in a circle.
$A$ and $B$ can be seated together in $35$ ways, of which $5$ cross "row barriers" and are invalid,
thus $Pr = \dfrac{30}{\dbinom{35}{2}},\approx0.05042$
A: Ways to select 2 seats out of 35 seats (5 rows * 7 seats) = ${35 \choose 2}$
Ways for seats to be adjacent = 6*5 (each row has 6 possible pairs of adjacent seats)
Hence, the required probability is 
$$\frac{6 \cdot 5}{35 \choose 2} = \frac{6 \cdot 5 \cdot 2}{35 \cdot 34} \simeq 0.05042 $$
