Suppose we have 4 couples and 8 seats. The seats are oriented such that it is a row. The people take their seats in a random fashion.
What is the probability that 2 couples sit next to each other?
My work: Let us label the people as $A_1, A_2, A_3, A_4, B_1, B_2, B_3, B_4$. $A_i$ and $B_i$ are couples. Then, we want to know the probability of $P(C_i \cap C_j), i \neq j$ and $C_i$ means couple $i$ are seated next to each other.
Then, suppose we fix $C_i$ and $C_j$. Then, the row arrangement could look something like: $\{ C_i, C_j, a,b,c,d\}$, where $a,b,c,d$ are the remaining 4 people. Then, there are $\binom{6}{2}$ ways of picking positions for $C_i$ and $C_j$, $2!$ ways of arranging the elements in $C_i$, $2!$ ways of arranging the elements in $C_j$, $2!$ ways of arranging $C_i$ and $C_j$ relative to each other, and $4!$ ways of arranging the remaining 4 people. There are $8!$ ways to arrange the 8 people in a row. Also, there are $\binom{4}{2}$ ways to pick 2 couples out of 4 couples.
Putting all this together, $ P(C_i \cap C_j) = \frac{\binom{6}{2} \binom{4}{2} 2! 2! 2! 4! }{8!}$
I am curious because this differs from the solution given here: cached solution from google
Note that on the bottom of page 5, the difference is the google solution is missing a $\binom{6}{2}$ term, which means we can choose 2 positions for $C_i$ and $C_j$ out of the available 6 positions. (Please observe the quantity $p_2$ in the linked document). The factor $\binom{4}{2}$ is accounted for later on the top of page 6, where the author sums up all the probabilities corresponding to possible indices $i$ and $j$.
Thanks.