permutations

apex231 wrote:One single person and two couples are to be 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

OA D

I solved like this -
Total ways to arrange = 5! = 120

Lets two couples be aa and bb.
Single person be c.

aabbc

Total ways that couples are together = 3! * 2 * 2 = 24 (multiply by 2 for same couple swapping places)

Total ways that couples are NOT together = 120 - 24 = 96

P = 96/120 = 4/5. But this is not the answer. where am i going wrong?

First let us find the probability of at least one of the couple being sitting together.
One single person and two couples implies in all there are 5 people, so no. of ways of seating these 5 people = 5! = 120 ways
Let us consider one couple as 1 person, then no. of ways of seating them = 4! ways and 1 couple means 2 people, who can be seated in 2! ways. So, no. of ways of seating them = 4! * 2! = 48 ways
Similarly, the other couple can be seated in 48 ways.
Hence, total no. of ways in which the 2 couples sit together = 3! * 2! * 2! = 24 ways
Now, the no. of ways so that at least one of the couple sit together = 48 + 48 - 24 = 72
Probability so that at least one of the couple sit together = 72/120 = 3/5

Therefore, probability that neither of the couples sits together in adjacent chairs = 1 - 3/5 = [spoiler]2/5[/spoiler]

The correct answer is D.