Probability

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Probability

by hey_thr67 » Tue Jun 19, 2012 2:17 am
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
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by GMATGuruNY » Tue Jun 19, 2012 3:08 am
hey_thr67 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
Let the 5 people consist of 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)

The big idea with overlapping groups is to SUBTRACT THE OVERLAP.
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.
The result is that the OVERLAP -- the arrangements in which both AB and CD sit together -- will be counted twice.
Thus, we SUBTRACT THE OVERLAP so that it is not double-counted.

Total arrangements = 5! = 120

AB together:
Here, we're arranging the 4 elements AB, C, D, and E.
Number of ways to arrangement 4 elements = 4! = 24.
Since AB can be reversed to BA, we multiply by 2:
2*24 = 48.

CD together:
Using the same reasoning used for AB, the number of arrangements here = 48.

AB and CD together:
Here, we're arranging the 3 elements AB, CD, and E.
Number of ways to arrange 3 elements = 3! = 6.
Since AB can be reversed, CD can be reversed, and both AB and CD can be reversed -- yielding a total of 4 variations -- we multiply by 4:
4*6 = 24.

Plugging these results 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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by dhonu121 » Tue Jun 19, 2012 10:06 am
Hi Mitch,
what is wrong if I do it the following way.
Lets group the couples AB as one element and CD as another element.
Then number of ways of arranging AB,CD and E, is 3!*2!*2!.
Total number of ways = 5!
Hence probability that all couples sit together is P = 3!*2!*2!/5!.
Thus probability that couples do not sit together will be 1-P.

The answer is coming out wrong. Would you please find the fault here ?

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