Problem Solving

Hi All!

Can anyone please provide a way to do this problem efficiently/quickly? What would be the most common approach to a problem such as this on the actual exam, if it were to be within the last 7 questions (with 12 mins to spare)?

Question:

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?

Possible Answer choices:
1. 1/5
2. ¼
3. 3/8
4. 2/5
5. ½

Thanks!

Dont know about strategy but you solve this one in less than 90 seconds

Ways in which the first couple can sit together = 2*4! (1 couple is considered one unit)
Ways for second couple = 2*4!
These cases include an extra case of both couples sitting together
Ways in which both couple are seated together = 2*2*3! = 4! (2 couples considered as 2 units- so each couple can be arrange between themselves in 2 ways and the 3 units in 3! Ways)
Thus total ways in which at least one couple is seated together = 2*4! + 2*4! - 4! = 3*4!
Total ways to arrange the 5 ppl = 5!
Thus, prob of at least one couple seated together = 3*4! / 5! = 3/5
Thus prob of none seated together = 1 - 3/5 = 2/5


Then you will have 10 mins 30 secs to solve rest 6 questions...

Is there another way to solve this problem? It is still a bit unclear to me. Thanks!