A, B, C, D, E and F are six friends attending a movie together. A and B want to sit together, C and D want to sit together, and E and F want to sit together. Unfortunately, the usher randomly assigns the six friends together in a row. What is the probability that all six friends sit next to whom they want?

Consider AB CD EF cases clubbed together
So the 6 choices are reduced to 3 choices.
Arranging the 3 cases can be done in 3! ways.
But AB can also be arranged as BA, CD by DC and EF by FE, ie 2x2x2=8 more possible cases.
Thus total possible cases come out to be 8x3!

Probability of desired event = No. of cases of desired event / Total No. of cases
= 8 x 3! / 6!
= 8 x 6 / 720
= 1/15

A, BB, CC, DD, EE and FF are six friends attending a movie together. AA and BB want to sit together, CC and DD want to sit together, and EE and FF want to sit together. Unfortunately, the usher randomly assigns the six friends together in a row. What is the probability that all six friends sit next to whom they want?

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