I attacked the question in the following way.
Totally there are 6! ways of seating arrangements possible which goes into the denominator.
Now in the numerator, we are told to find the prob of them sitting next to each other.
So AB can be seated together in 3 different ways, ie; AB in the first 2 seats, AB in the middle 2 and AB in the last 2. This totally gives us 3 ways of arranging. Also, note that AB can BA in those positions too. So there are 6 ways in which they can be seated together.
The same applies to the other two pairs CD and EF.
So in the numerator, we will have 6 * 6* 6 / 6! which gives us 3/20
which is not in the answer choice.
Am I doing anything wrong? Help me out & thanks in advance 