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

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Manager
Joined: 28 Jun 2004
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Two couples and one single person are seated at random in a [#permalink]

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22 May 2005, 13:06
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Two couples and one single person are seated at random in a row of 5 chairs. What is the probability that neither of the couples sits together in adjacent chairs?
Director
Joined: 18 Feb 2005
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22 May 2005, 13:19
Let me know if 13/24 is correct. I would explain then
Manager
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22 May 2005, 13:24
Nope, the OA is 2/5. And frankly, I cannot understand their explanation. Its a 25 line explanation.
Director
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22 May 2005, 13:27
This was discussed before...Heres the link

http://www.gmatclub.com/phpbb/viewtopic ... 9976#89976
Director
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22 May 2005, 21:51
Can some one let me know the gap in this process?

Total outcomes = 5! =120

Lets says that we have the 2 couple as 2 blocks occupying 4 seats.

SO we have 2 blocks( each block has a couple) and 1 individual.

So they can be arranges in 3!.2!.2! (Since the couples can shuffle between themselves)

So if they sit together it would be 24/120 = 1/5 and probablity of not sitting together is 4/5....

Whats the flaw in this logic? Please explain
SVP
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22 May 2005, 22:08
What if one of the couples sits together and the other doesn't?
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
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Director
Joined: 18 Apr 2005
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Location: Canuckland
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22 May 2005, 22:08
gmat2me2 wrote:
Can some one let me know the gap in this process?

Total outcomes = 5! =120

Lets says that we have the 2 couple as 2 blocks occupying 4 seats.

SO we have 2 blocks( each block has a couple) and 1 individual.

So they can be arranges in 3!.2!.2! (Since the couples can shuffle between themselves)

So if they sit together it would be 24/120 = 1/5 and probablity of not sitting together is 4/5....

Whats the flaw in this logic? Please explain

What if one couple sits together and one doesn't?
SVP
Joined: 03 Jan 2005
Posts: 2236
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Kudos [?]: 342 [0], given: 0

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22 May 2005, 22:10
Great minds think alike don't they?
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Director
Joined: 18 Apr 2005
Posts: 547
Location: Canuckland
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22 May 2005, 22:10
Just to clarify, HongHu and I aren't the same person. Just happened to post at the same time
Director
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22 May 2005, 22:13
HongHu wrote:
What if one of the couples sits together and the other doesn't?

But we were asked in the question how many ways the 2 couples cannot sit together....So I subtracted 1 - (Probablity of 2 couples sitting together)....

So you mean to say that 1 couple sit together but the other aint...Isnt it ?

So that could be 4!*2 ...Is this correct?
Director
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22 May 2005, 22:15
sparky wrote:
Just to clarify, HongHu and I aren't the same person. Just happened to post at the same time :)

Boy you guys scared me enough Its 12:15 AM here I am going to bed :-)
Director
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23 May 2005, 21:47
gmat2me2 wrote:
HongHu wrote:
What if one of the couples sits together and the other doesn't?

But we were asked in the question how many ways the 2 couples cannot sit together....So I subtracted 1 - (Probablity of 2 couples sitting together)....

So you mean to say that 1 couple sit together but the other aint...Isnt it ?

So that could be 4!*2 ...Is this correct?

So the each of the 2 couples can sit together in 4!*2! ways = 48 ways

Since there are 2 couple = 96 ways.

But this included both the couples sitting toghether = 3!*2!*2! = 24 (I was confused at this point but now I think I understand the question)

96-24 = 72 ways only one couple can sit togther

probability of Neither of couple sitting together is= 72/120 = 3/5
23 May 2005, 21:47
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# Two couples and one single person are seated at random in a

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