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TWO couples and a single person are to be seated on 5 chairs

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Manager
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Joined: 16 Oct 2008
Posts: 54

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TWO couples and a single person are to be seated on 5 chairs [#permalink]

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New post 28 Oct 2008, 05:53
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

4. TWO couples and a single person are to be seated on 5 chairs such that no couple is seated next to each other. What is the probability of the above??

Kudos [?]: 2 [0], given: 0

Director
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Joined: 14 Aug 2007
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Re: Please help # 4/4 [#permalink]

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New post 28 Oct 2008, 06:35
Natasha0123 wrote:
4. TWO couples and a single person are to be seated on 5 chairs such that no couple is seated next to each other. What is the probability of the above??


Total number of ways=5!
Possible number of ways =
Consider single person alone for time being and move him in the 5 chairs.
Single person in
1st chair = 1 * 4 * 2 * 1 * 1
2nd chair = 4 * 1 * 2 * 1 * 1
3rd chair = 4 * 2 * 1 * 2 * 1
4th chair = 1 * 1 * 2 * 1 * 4
5th chair = 1 * 1 * 2 * 4 * 1
Adds up to 48

prob = 48/5!

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Manager
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Joined: 16 Oct 2008
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Kudos [?]: 2 [0], given: 0

Re: Please help # 4/4 [#permalink]

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New post 28 Oct 2008, 07:01
This is the solution, but I don't understand...

Ways in which the first couple can sit together = 2*4! (1 couple is considered one unit)
Ways for second couple = 2*4!
These cases include an extra case of both couples sitting together
Ways in which both couple are seated together = 2*2*3! = 4! (2 couples considered as 2 units- so each couple can be arrange between themselves in 2 ways and the 3 units in 3! Ways)
Thus total ways in which at least one couple is seated together = 2*4! + 2*4! - 4! = 3*4!
>>> really confuse

Total ways to arrange the 5 ppl = 5!
Thus, prob of at least one couple seated together = 3*4! / 5! = 3/5

Thus prob of none seated together = 1 - 3/5 = 2/5

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VP
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Re: Please help # 4/4 [#permalink]

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New post 28 Oct 2008, 13:12
Natasha0123 wrote:
4. TWO couples and a single person are to be seated on 5 chairs such that no couple is seated next to each other. What is the probability of the above??


Initially I read it to be the number of such arrangements

Any ways 5 people in 5 chairs in 5! ways.

Now we need to find arrangements such that no couple is together. lets find together combinations. bundle each couple as one unit. means 3 people in 3! ways. Each couple can be arranged in 2 different ways and there are 2 such couples so it is 2 X 2 X 3! = 24

Required = 120 - 24 = 96

Probability = 96/120 = 4/5

Wait! My answer is wrong :x

Kudos [?]: 437 [0], given: 1

Re: Please help # 4/4   [#permalink] 28 Oct 2008, 13:12
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TWO couples and a single person are to be seated on 5 chairs

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