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Two couples and one single person are seated at random in a

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Two couples and one single person are seated at random in a [#permalink]  10 Oct 2004, 14:31
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?

a) 1/5
b) 1/4
c) 3/8
d) 2/5
e) 1/2
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[#permalink]  10 Oct 2004, 22:03
D: 2/5.

3 persons can sit on 5 chairs in 5P3 = 20 ways. (Total possible ways)

now lets assume the 2 couples to be a single entity and then there wud be effectively 4 chairs :so 2 people can sit on 4 chairs in 4P2 = 12 ways.

So the no. of ways in which the couples are not sitted adjacent is 20-12 = 8.

Hence the reqd. probability is 8/20 = 2/5.
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5 or 3? [#permalink]  11 Oct 2004, 01:38
I thought two couples mean total 4 person. So, there are 5 chairs and 5 people. total way = 5p5=120.

we take couples as same entity and both are sitting together, we got:

No. of ways = 3p3*2*2=24

If one of them is sitting together, we got:

No. of ways = 4p4*2=48

Probability = [120- (24+48)]/120=2/5.

Shishir
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[#permalink]  11 Oct 2004, 19:18
I agree it should be 2 couples = 4 people, thus there are 5 people and 5 chairs. I guess I'm not understanding why it can work with 5P3...

Anyway, here's my answer

5! = 120 total ways 5 people can sit

Both couples sitting adjacent = Group two couples together and multiply by each different way a couple can sit adjacent = 3! * 2 * 2 = 24

First couple sitting adjacent but not both sitting adjacent = First couple together and subtract both sitting together = 4! - (3! * 2 * 2) = 24

Second couple sitting adjacent but not both sitting adjacent = Second couple together and subtract both sitting together = 4! - (3! * 2 * 2) = 24

Probability = (120 - 24 - 24 - 24)/120 = 48/120 = 2/5
CIO
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[#permalink]  12 Oct 2004, 06:57
I like all the ways given, but not sure someone without a complete comprehension of probability and permutations would get it. I have a way that's a little more cumbersome, but it works:

There are 5 spaces to fill up:
_ _ _ _ _

The denominator here will be 120, since that's the total number of ways 5 people could sit in 5 seats.

The number of permutations for the numerator would be this:

the single person could be in any of those 5 spaces:

S _ _ _ _
_ S _ _ _
_ _ S _ _
_ _ _ S _
_ _ _ _ S

Now go through and think logically about how many people could fill each of the other spaces:

S 4 2 1 1 = 8
4 S 2 1 1 = 8
4 2 S 2 1 = 16
4 2 1 S 1 = 8
4 2 1 1 S = 8

So the total number of possibilities is 48/120 = 2/5

I know this seems longer, but some people (like myself) think more clearly this way, rather than trying to conceptualize the bouncing around of couples in a permutations world. Since there are only 120 total possibilities, it's not impossible to see this method being worthwhile on the test, even if it adds an extra minute.
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[#permalink]  14 Oct 2004, 04:23
OA is 2/5. Great Job and Thanks!!!
[#permalink] 14 Oct 2004, 04:23
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