Chaitanya_1986 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 is D[/spoiler]

We could treat this as an overlapping groups problem:

Total = G1 + G2 - Both + Neither

Let's say that we have couple AB, couple CD, and lonely person E.

Total arrangements = (arrangements with AB together) + (arrangements with CD together) - (arrangements with both AB and CD together) + (arrangements with neither AB nor CD together)

We need to subtract from our total the arrangements in which both AB and CD sit together because when we count the arrangements in which AB sit together, among those arrangements will be some in which CD also sit together. When we count the arrangements in which CD sit together, among those arrangements will be some in which AB also sit together. So the arrangements in which both AB and CD sit together -- the overlap -- will have been counted twice.

Total arrangements = 5! = 120

Total arrangements with AB together means we're arranging the 4 elements AB, C, D, and E = 4! = 24. Now we multiply by 2 since AB can be reversed to BA: 2*24 = 48.

Total arrangements with CD together means we're arranging the 4 elements CD, A, B and E = 4! = 24. Now we multiply by 2 since CD can be reversed to DC: 2*24 = 48.

Total arrangements with AB and CD together means we're arranging the 3 elements AB, CD, and E = 3! = 6. Now we multiply by 4 because AB can be reversed, CD can be reversed, and both AB and CD can be reversed: 4*6 = 24.

Plugging into the overlapping groups formula, we get:

120 = 48 + 48 - 24 + N
120 = 72 + N
N = 48

So P(neither couple sitting together) = 48/120 = 2/5.

Thanks Anurag and Mitch!!!!

@All

Let me make it SIMPLER

Probability of Neither of the couple sitting together = 1-(Probabaility of at least ONE couple sitting together+Probability of BOTH the couples sitting together)

a) Probability of AT LEAST ONE couple sitting together = (5-2+1)!*2!/5!= 2/5

b) Probability of BOTH the couples sitting together = (5-2-2+1+1)!*2!*2!/5! = 1/5

so ,Probability of Neither of the couple sitting together = 1-(1/5+2/5) = 2/5

And the answer choice is (D)

Does the below make sense -- though it yields the same answer I am not sure whether my logic is correct.

p(neither together) = 1 - p(both with their mates)

solving positive event.

consider couples as unit -- AA BB S
there are 3! ways of arranging the above
2!^2 ways between the couple
3 ways in which the single guy can fit in

hence desired = 3! X 2!^2 X 3 = 72
total possibilities = 5!

p(+ve) = 72/120 = 3/5

hence p(-ve) = 1 - p(+ve) = 1-3/5 = 2/5.

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another way i got this is -
consider unit method, there are 3 units out of 5

therefore 3/5 for p(altogether), and p(neither) = 2/5 [1-3/5]

Really a gud question. The explanation provided by Anurag is the method to be used when multiple "COUPLES" are involved.