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.
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