Seat friends at a dinner table I have the following question:
At a squared table I seat $8$ friends of mine.
At each side, I place two friends.
Always two friends know each other.
What's the probability that no friend knows it's sided, neighbor?
I tried to solve it by:


*

*Calculating all different seating possibilities: $4!$ $\Rightarrow 24$

*And writing down all options on paper where no side neighbors know the other side neighbor. For that, I got 9 options. 


So I guess my final result is  $$\frac{9}{4!}\Rightarrow 0.375$$
$37.5\%$ that none of the side neighbors knows the other side neighbor.
Is this correct? How would I calculate the count of all possible seating options where no side neighbor knows each other?
Thanks
 A: A slight improvement on user8734617's method:
Exactly $\frac{1}{7}$ of the possible arrangements have a fixed pair $A_1,B_1$ seated together. To see this, given any arrangement with the pair seated together, fix the position of $A_1$ and rotate the other $7$ around the table, giving $6$ other arrangements with $A_1,B_1$ not seated together. This method produces all possible arrangements around the table, so each arrangement with $A_1,B_1$ seated together corresponds to $6$ other arrangements, meaning $A_1,B_1$ sit together with probability $\frac{1}{7}$.
By the same method, of those arrangements with a fixed pair $A_1,B_1$ seated together, $\frac{1}{5}$ have another fixed pair $A_2,B_2$ also seated together, so two fixed pairs sit together with probability $\frac{1}{7 \cdot 5}$; similarly, three fixed pairs sit together with probability $\frac{1}{7 \cdot 5 \cdot 3}$ and four with probability $\frac{1}{7\cdot5\cdot3\cdot1}$.
Then we apply inclusion-exclusion: our answer is $$\begin{align}
&1 - \binom{4}{1}\cdot\frac{1}{7} + \binom{4}{2}\cdot\frac{1}{7 \cdot 5} - \binom{4}{3}\cdot\frac{1}{7 \cdot 5 \cdot 3} + \binom{4}{4}\cdot\frac{1}{7\cdot5\cdot3\cdot1} \\
&= 1 - \frac{4}{7} + \frac{6}{7\cdot5} - \frac{4}{7 \cdot 5 \cdot 3} + \frac{1}{7 \cdot 5 \cdot 3} \\
&= \frac{3}{7} + \frac{6}{7\cdot5} - \frac{3}{7 \cdot 5 \cdot 3} \\
&= \frac{3}{7} + \frac{5}{7\cdot5} \\ &= \frac{4}{7}
\end{align}$$
A: Let's label the friends $A_1, A_2, A_3, A_4, B_1, B_2, B_3, B_4$ where $A_i$ knows $B_i$.
Using inclusion-exclusion formula: the number of ways to put them around the table so that none of $A_i$ sits next to the corresponding $B_i$ is:
$$\sum_{k=0}^{4}(-1)^k{4\choose k}^22^kk!(8-2k)!$$
Each term in the sum is the sum over all the sets of $k$ (out of 4) pairs of friends presuming they are together (but not presuming that no others are together). The factor $-1$ comes from the inclusion-exclusion formula itself, one of ${4\choose k}$ comes from the fact that there are that many sets of $k$ pairs, the other ${4\choose k}$ is the number of ways to put those $k$ pairs around the table, $k!$ is to allow for permutations of those $k$ pairs, $2^k$ comes from permuting each pair once they are already seated, and $(8-2k)!$ comes from putting the rest of the people ($8-2k$) around the table arbitrarily.
I am not sure whether there is a shortcut to calculate the total number, but the formula above is short enough to do a manual calculation:
$$8!-4^22^11!6!+6^22^22!4!-4^22^33!2!+1^22^44!0!=40320-23040+6912-1536+384=23040$$
so the probability that no two friends sit together is $\frac{23040}{8!}=\frac{4}{7}$.
I have to say that, now I've got this result, and it looks very simple, I suspect there is an alternative and much simpler calculation. If there is, and if I can get to it, I will update this answer (unless someone else beats me to it).
A: I know that it is now done and dusted, but here is a direct method avoiding PIE that may appeal.
We needn't bother about numbering $A_1,A_2,...$ or for that matter, the seat numbers.
Just treat them as balls of $4$  colors and follow the chain of pairings to get pairs in unlike colors.
Take any ball, and start pairing. If no pairs are to be similar, $Pr = \frac67\cdot\frac45\cdot\frac23 = \frac{48}{105}$
Alternatively, two pairs each can be alike, e.g. $AC\;\; AC\;\; BD\;\; BD,\; Pr = (\frac67\cdot\frac15)(\frac23) =\frac{12}{105}$
Adding up, ans $= \frac{60}{105} = \frac47$ 

Added explanation
Suppose we first pick an $A$. Out of the $7$ other balls, we need to avoid the other $A$, so $\frac 67$, and let's say we paired it with  $C$. If we want no pairs to be similar, the other $C$ can't be paired with the "free" $A$, so $\frac45$ and so on.
On the other hand, for like pairs, having got $AC$, the other $C$ has to be paired with $A$, thus $\frac67\cdot\frac15$, and now we have $B\;\;B\;\;D\;\;D\;$ left to pair similarly. 
