If $3$ married couples are seated randomly in a row, then find the probability that exactly $2$ couples can sit together.

Question

If $3$ married couples are seated randomly in a row, then find the probability that exactly $2$ couples can sit together.

My attempt is like this: $$\frac{4!(2!)^2}{6!}$$ My understanding: The probability of $3$ couples ($6$ people) sit in a row is 6!. Then, if $2$ couples must sit together, there will be $4$ units in a row (i.e. $4!$). But inside those $2$ couples, they can still exchange the seat. Therefore, it will be $(2!)^2$.

However, the answer is $$3C1 \cdot \left[\frac{4!(2!)^2}{6!} - \frac{3!(2!)^3}{6!}\right]$$

I don't understand why my attempt is incorrect.

Please Help. Thanks!

Answer 1

Your numerator counts the number of arrangements in which couples $A$ and $B$ are together. To find the number of arrangements in which couples $A$ and $B$ are together and couple $C$ are not, we must subtract the number of arrangements in which they are all together, which is $(3!)(2!)^3$.

So your expression, minus $(3!)(2!)^3$, counts the number of ways couple $A$ and $B$ are together, but couple $C$ are not. Finally, we need to multiply by $3$, for the couple unlucky (lucky?) enough to be separated can be chosen in $3$ ways.