Two weeks ago I came up with the following probability problem (maybe a similar one already exists):
A plane has $60$ seats, disposed by groups of $3$ seats, $2$ groups per row ($10$ rows):
$$ \text{cockpit} \\ S_{1}S_{2}S_{3}\text{ }\text{ }\text{ }\text{ }\text{ }S_{4}S_{5}S_{6} \\ S_{7}S_{8}S_{9}\text{ }\text{ }\text{ }\text{ }\text{ }S_{10}S_{11}S_{12} \\ ... \\ S_{55}S_{56}S_{57}\text{ }\text{ }\text{ }\text{ }\text{ }S_{58}S_{59}S_{60} \\ \text{tail of the plane} $$
All seats were booked and seats are assigned randomly to each person with an equal probability.
What is the probability that person $p_{1}$ and person $p_{2}$ will sit next to each other?
My solution to the problem was the following:
There are $60!$ possible outcomes, let $p_{n}$ be a person sitting on a seat:
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ } 1)\text{ Starting from seat number 1: } p_{1}p_{2}p_{3} \text{ }\text{ }\text{ } p_{4}p_{5}p_{6} \text{ }\text{ }\text{ } ... \text{ }\text{ }\text{ } p_{58}p_{59}p_{60}\text{ (until seat number 60)}$
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ } 2)\text{ Starting from seat number 1: } \text{ } p_{1}p_{3}p_{2} \text{ }\text{ }\text{ } p_{4}p_{5}p_{6} \text{ }\text{ }\text{ } ... \text{ }\text{ }\text{ } p_{58}p_{59}p_{60} \text{ (until seat number 60)}$
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ } ...$
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ } 60!)\text{ Starting from seat number 1: }\text{ } p_{60}p_{59}p_{58} \text{ }\text{ }\text{ } p_{57}p_{56}p_{55} \text{ }\text{ }\text{ } ... \text{ }\text{ }\text{ } p_{3}p_{2}p_{1}\text{ (until seat number 60)}$
Then the outcomes where the event $p_{1}\text{ sits next to }p_{2}$ occurs are the following (note that seats are grouped by 3):
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ }1) \text{ } \color{red}{p_{1}p_{2}}p_{3} \text{ }\text{ }\text{ } p_{4}p_{5}p_{6} \text{ }\text{ }\text{ } ... \text{ }\text{ }\text{ } p_{58}p_{59}p_{60}$
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ }2) \text{ } \color{red}{p_{2}p_{1}}p_{3} \text{ }\text{ }\text{ } p_{4}p_{5}p_{6} \text{ }\text{ }\text{ } ... \text{ }\text{ }\text{ } p_{58}p_{59}p_{60}$
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ }3) \text{ } p_{3}\color{red}{p_{1}p_{2}} \text{ }\text{ }\text{ } p_{4}p_{5}p_{6} \text{ }\text{ }\text{ } ... \text{ }\text{ }\text{ } p_{58}p_{59}p_{60}$
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ }4) \text{ } p_{3}\color{red}{p_{2}p_{1}} \text{ }\text{ }\text{ } p_{4}p_{5}p_{6} \text{ }\text{ }\text{ } ... \text{ }\text{ }\text{ } p_{58}p_{59}p_{60}$
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ }5) \text{ } p_{3}p_{4}p_{5} \text{ }\text{ }\text{ } \color{red}{p_{1}p_{2}}p_{6} \text{ }\text{ }\text{ } ... \text{ }\text{ }\text{ } p_{58}p_{59}p_{60}$
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ }6) \text{ } p_{3}p_{4}p_{5} \text{ }\text{ }\text{ } \color{red}{p_{2}p_{1}}p_{6} \text{ }\text{ }\text{ } ... \text{ }\text{ }\text{ } p_{58}p_{59}p_{60}$
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ }...$
Since for each group of 3 seats there are 4 possible cases when $p1$ and $p2$ sit next to each other, e.g.:
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \color{red}{p_{1}p_{2}}p_{3}$
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \color{red}{p_{2}p_{1}}p_{3}$
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } p_{3}\color{red}{p_{1}p_{2}}$
$\text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } p_{3}\color{red}{p_{2}p_{1}}$
and since there are $20$ groups of seats in the plane, each group with $3$ seats, I would say that the outcomes where $p_{1}$ and $p_{2}$ sit next to each other are $20 \cdot 4 \cdot (58!) = 80 \cdot 58!$ (20 groups times the 4 possible cases per group for which the statement $p_{1} \text{ sits next to } p_{2}$ holds true times the number of permutations of the remaining $60 - 2 = 58$ persons).
Therefore my answer would be:
$P(p_{1} \text{ sits next to } p_{2}) = \frac{\frac{60}{3}\cdot 4 \cdot 58!}{60!} = \frac{20 \cdot 4 \cdot (58!)}{60!} = \frac{20 \cdot 4 \cdot (58!)}{60 \cdot 59 \cdot (58!)} = \frac{80}{60 \cdot 59} = 0.02259887005 = 2.259887005\text{%}$
Are my reasoning and solution correct?
Thank you for the attention.