TWO couples and a single person are to be seated on 5 chairs such that no couple is seated next to each other. What is the probability of the above??
Soln:
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
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The above solution comes from the list of 100 hard GMAT problems posted. I have a problem with the part of acccounting for the extra case in which both couples are seated together. 2*2*3! treats the couples as units, making a total of 3 units. For example let A be one couple, B be another couple and S be the single individual. We are trying to eliminate the possibilty of having ABS, or SAB. But 3! also involves ASB, in which case the two couples are not seated together. Why should we eliminate such a case? It sounds more logical to treat both couples as single in which case we will have 2*2*2!. So the only possibilities are ABS SAB where couples are free to shift between themselves as captured by 2*2.

Any response to the above?