Three couples sit at random in a line of six seats, probability that no couple sits together? If three married couples (so 6 people) sit in a row of six seats at random, what is the probability that no couples sit together?
Another way to think about it (couples are AB, CD, and EF)


There are $6!$ ways of seating $6$ people ${H_1, W_1, H_2, W_2, H_3, W_3}$ in a row of six seats. 
Let's now treat each couple as a unit $U$, so we've got $$U_1 = H_1, W_1$$ $$U_2 = H_2, W_2$$ $$U_3 = H_3, W_3$$
Now, there are $2*4!$ ways of seating $3$ units of two people in row of $3$ seats.
It follows that there are $6! - 2*4! = 672$ ways in which no couple sit nex to each other. 
Hence the probability is $$\frac{672}{6!} = \frac{14}{15}$$
This is how I, personally, would go about it. 
