<s>[INST] If $3$ couples are randomly seated, what is they probability that at least $1$ couple sits together? My attempt: 
$$A = \text{couple 1 sits together}$$
$$B = \text{couple 2 sits together}$$
$$C = \text{couple 3 sits together}$$
Solving for
$$P(A \cup B \cup C) = P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C)-P(B \cap C)+P(A \cap B \cap C)$$
$$P(A)=P(B)=P(C) \quad\text{and}\quad P(A \cap B)=P(A \cap C)=P(B \cap C)\text{, so}$$
$$P(A \cup B \cup C) = 3P(A)-3P(A \cap B)+P(A \cap B \cap C)$$
\begin{align*}
P(A) = \frac{5!}{6!}\quad&\text{if 1 couple acts as a unit}\\
P(A \cap B) = \frac{4!}{6!}\quad&\text{if 2 couples act as single units}\\
P(A \cap B \cap C) = \frac{3!}{6!}\quad&\text{if all couples act as single units}
\end{align*}
 [/INST] You have the right idea, but you have not taken into account the internal arrangements of the couples that sit together.  A couple can sit together in $2!$ ways.  Hence, you should have 
\begin{align*}
\Pr(A) & = \frac{5!2!}{6!}\\
\Pr(A \cap B) & = \frac{4!2!^2}{6!}\\
\Pr(A \cap B \cap C) & = \frac{3!2!^3}{6!}
\end{align*}
$\Pr(A)$:  There is a couple that sits together and four other individuals, so we have five objects to arrange, which can be done in $5!$ ways.  The couple can be arranged internally in $2!$ ways.  Hence, the number of favorable arrangements is $5!2!$.  
$\Pr(B)$:  There are two couples and two other individuals, so we have four objects to arrange, which can be done in $4!$ ways.  Each couple can be arranged internally in $2!$ ways.  Hence, there are $4!2!^2$ favorable cases.  
$\Pr(C)$:  There are three couples to arrange, which can be done in $3!$ ways.  Each couple can be arranged internally in $2!$ ways.  Hence, there are $3!2!^3$ favorable cases. 
With these corrections, we obtain
$$\Pr(A \cup B \cup C) = \binom{3}{1}\frac{5!2!}{6!} - \binom{3}{2}\frac{4!2!^2}{6!} + \binom{3}{3}\frac{3!2!^3}{6!}$$
</s>