7 seats in a row, probability of the married couple sitting together So I've figured out that there are $7 \choose 2$ = $21$ possible seat pairings for the couple, and 6 of them result in sitting together, so the probability of them sitting together is $\frac 27$. 
However, why doesn't order matter here? Isn't a seat pairing with husband/wife in seat 1/2 respectively, different than them being in seats 2/1 respectively? 
Also, why don't we need to account for all the ways the other 5 people could arrange themselves?  
 A: If order matters then then are $7\times 6 = 42$ ways of sitting the wife then the husband.  There are then $5! = 120$ ways of sitting the other people, making $7!=5040$ ways of seating everybody
Of these, there $2 \times 6=12$ ways of seating the wife and husband together when order matters, or $2 \times 6!=1440$ ways if you also count the other people
All these approaches give the same answer, as $\dfrac{6}{21}=\dfrac{12}{42}=\dfrac{1440}{5040}=\dfrac{2}{7}$  
The important thing is keep a consistent method of counting equally probable results for the numerator and denominator  
A: Say $\Omega_1$ is the probability space consisting of the possible orderings of distinct people and each ordering has the same probability. What you did is to map this space to $\Omega_2$ where we forget the identity of the individuals and only distinguish between the married pair and the rest. So for instance $\Omega_2$ consists of all the orderings of 2 red and 5 blue marbles, if you will.
Say person 1 and 2 are married, then 
$$ \phi : \Omega_1 \to \Omega_2 $$
$$ (5,3,2,6,4,1,7) \mapsto (B,B,R,B,B,R,B) $$
Via $\phi$, the probability measure on $\Omega_1$ induces a measure on $\Omega_2$; but it turns out this measure is the SAME as the uniform distribution on $\Omega_2$.
This is because each preimage $\phi^{-1}((B,B,R,B,B,R;B))$ has the same size, namely $2\cdot 5 = 10$, and therefore the same probability.
Anti-example: Due to unfortunate circumstances the last two seats in the row explode. What is the probability of the couple sitting together now?
One guess is: $\frac{4}{\binom{5}{0} + \binom{5}{1} + \binom{5}{2}}$, i.e. either one or two of the spouses got blown up, or both are alive with 5 seats left. But this is not the right answer: Let $\Omega_3$ be the set of possible orderings of red and blue marbles after the explosion.
The map $\Omega_2 \mapsto \Omega_3, (x_1,...,x_7) \mapsto (x_1,...,x_6)$ has preimages of cardinality 1 or 2, so the above answer is not correct.
