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?
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Took me some time to answer this question from my own logic. Just want to know if my approach is right. The solution video has a very simple way of solving this but I want to check if my thinking and way of solving is correct and valid.
Here it is—
Since there are 6 friends, the total number of combinations possible = 6! = 720
The total number of combinations possilbe if they all want to sit with their desired partners —
2 CHOICES and 2 CHOICES and 2 CHOICES = 2 x 2 x 2 = 8
(AB or BA) (CD or DC) (EF or FE)
Since there can be 6 ways of arranging these three groups,
8 x 6 = 48
Therefore 48/720
after simplifying = 1/15
which is the correct answer.
Is this the right way to approach this question?