probablities anyone ?

[winna.7]

Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs? A) 1/5 B) 1/4 C) 3/8 D) 2/5 E) 1/2

I came across this one somewhere and think i solved it.My answer was 11/20 .Butttt... ...
My answer did not match any of the choices. Could anyone please tell me if i have gone wrong some place here? this was i went about it :


Consider the 5 people to be AABBC .
The probability that both couples are seated together is 3P3 (it is as good as considering each couple as one member)

The probability that only one of the couple is together is 4P4 (it is as good as considering the coupled couple as one member)

Now, this holds for both the situations where AA are together and BB are together. So the total no of ways in which the seating can be done with at least one couple together is = 3P3+ 4P4 + 4P4 = 54

Now the total no of ways seating can be done = 5P5 = 120

The total no ways in which seating can be done such that neither of the couples are seated together is = 5P5 - (3P3+ 4P4 + 4P4)= 120-54= 66

So the probability that neither of the couples are together is 66/120 =
11/20 .

Am i wrong some where ?

[winna.7]

Oops ! posted it in the wrong section ! im sorry..! i have reposted this in the math section. Please ignore this one.