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9 people, including 3 couples, are to be seated in a row of

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9 people, including 3 couples, are to be seated in a row of [#permalink] New post 11 Apr 2005, 04:45
9 people, including 3 couples, are to be seated in a row of 9 chairs.


What is the probability that


a. None of the Couples are sitting together


b. Only one couple is sitting together


c. All the couples are sitting together
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 [#permalink] New post 11 Apr 2005, 05:57
I will try to solve C) All possible arrangements are 9!. When all couples are together 7x5x3x8x6 or 8x7x6x5x3/9! is 1/72
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 [#permalink] New post 11 Apr 2005, 07:06
BG wrote:
I will try to solve C) All possible arrangements are 9!. When all couples are together 7x5x3x8x6 or 8x7x6x5x3/9! is 1/72


Is that a 0.0138 chance that the couples will not be able to sit together?
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Re: Probability: 9 People, 3 couples and 9 chairs [#permalink] New post 11 Apr 2005, 12:10
mbassmbass04 wrote:
9 people, including 3 couples, are to be seated in a row of 9 chairs.


What is the probability that


a. None of the Couples are sitting together


b. Only one couple is sitting together


c. All the couples are sitting together


a) derives from c), so I'd solve c) first.

If all couples are sitting together, then permutations are 6! x 2 x 2 x 2
= 720 x 6 = 4320.

(Btw this means the couples sit together, not ALL the couples sit together. For example, if A, B are couples, C, D are couples, E, F are couples, and G H I are happily single, then AB are together, CD are together, and EF are together but does not necessarily mean that ABCDEF are together - I would post a solution for that later)

a) Number of ways in which none of the couples are sitting together = Total number of ways - ways in which all couples are sitting together
= 9! - 4320
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 [#permalink] New post 11 Apr 2005, 12:23
only one couple sitting together? (question b)

8!*2
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 [#permalink] New post 11 Apr 2005, 14:35
thearch wrote:
only one couple sitting together? (question b)

8!*2


:) theArch, think again. Aren't you forgetting something?

This case means you consider 1 couple as one entity and dont care about other couples. But this case includes cases when the other couples are together and separate (in any combination basically).

What say? :)
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 [#permalink] New post 12 Apr 2005, 00:00
you know, I learnt the existence of this kind of math here :-D , never studied it...
now I'll give it a try :war

8!*2-7!*2*2-6!*2*2*2

(At a first glance, I thought it was negative, then I checked it and it is 54720)

praise me if I'm right, don't kill me if I'm wrong :-D
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Re: Probability: 9 People, 3 couples and 9 chairs [#permalink] New post 12 Apr 2005, 02:22
Quote:
a) derives from c), so I'd solve c) first.

If all couples are sitting together, then permutations are 6! x 2 x 2 x 2
= 720 x 6 = 4320.

(Btw this means the couples sit together, not ALL the couples sit together. For example, if A, B are couples, C, D are couples, E, F are couples, and G H I are happily single, then AB are together, CD are together, and EF are together but does not necessarily mean that ABCDEF are together - I would post a solution for that later)

a) Number of ways in which none of the couples are sitting together = Total number of ways - ways in which all couples are sitting together
= 9! - 4320


kapslock, is your answer right ?

what if only one couple sits together or only two ? IMO you just calculated the comb when exactly ALL three sit together. correct if i am wrong...
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Re: Probability: 9 People, 3 couples and 9 chairs [#permalink] New post 12 Apr 2005, 15:35
christoph wrote:
Quote:
a) derives from c), so I'd solve c) first.

If all couples are sitting together, then permutations are 6! x 2 x 2 x 2
= 720 x 6 = 4320.

(Btw this means the couples sit together, not ALL the couples sit together. For example, if A, B are couples, C, D are couples, E, F are couples, and G H I are happily single, then AB are together, CD are together, and EF are together but does not necessarily mean that ABCDEF are together - I would post a solution for that later)

a) Number of ways in which none of the couples are sitting together = Total number of ways - ways in which all couples are sitting together
= 9! - 4320


kapslock, is your answer right ?

what if only one couple sits together or only two ? IMO you just calculated the comb when exactly ALL three sit together. correct if i am wrong...


Christoph,

You're right. I just calculated the combinations when exactly ALL three sit together, in case (c). But isn't this what was required to be done? I am copying and pasting the question statement here.


"9 people, including 3 couples, are to be seated in a row of 9 chairs.
What is the probability that
a. None of the Couples are sitting together
b. Only one couple is sitting together
c. All the couples are sitting together".

c directly leads to case a. I am not sure of b. Its a bit tougher than a and b. Probably I'd try it at home.

theArch, why don't you post your approach to the solution of (b) - saves precious brain fuel :)
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 [#permalink] New post 12 Apr 2005, 15:55
Guys please correct my approach if it is wrong
9! - 2*7! *2*2


Take 2 couples as 1 entity you get 7 arrangements to be made

The pair of couples could be arraged in 2 ways BETWEEN the 2 couples.

Total- 2 couples sitting together
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 [#permalink] New post 12 Apr 2005, 16:29
thearch wrote:
you know, I learnt the existence of this kind of math here :-D , never studied it...
now I'll give it a try :war

8!*2-7!*2*2-6!*2*2*2

(At a first glance, I thought it was negative, then I checked it and it is 54720)

praise me if I'm right, don't kill me if I'm wrong :-D


Awesome theArch.

I bow before you, for you're God.

That's what I meant, that if you consider 1 couple together, and considering ALL permutations of remaining people, you're including cases where other couples are together as well. So you very deftly exclude those cases as well, where those couples are together. And hey presto !!!

You rock theArch !!! :band
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 [#permalink] New post 13 Apr 2005, 00:49
Quote:
(Btw this means the couples sit together, not ALL the couples sit together. For example, if A, B are couples, C, D are couples, E, F are couples, and G H I are happily single, then AB are together, CD are together, and EF are together but does not necessarily mean that ABCDEF are together - I would post a solution for that later)


kapslock, i mentioned it, because in you statement above you said that ALL couples sit together doesnt mean that ALL sit together, but that AB and/or and/or CD and/or EF sit together. if you stick to this, your calc would be wrong.
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 [#permalink] New post 13 Apr 2005, 01:48
christoph wrote:
Quote:
(Btw this means the couples sit together, not ALL the couples sit together. For example, if A, B are couples, C, D are couples, E, F are couples, and G H I are happily single, then AB are together, CD are together, and EF are together but does not necessarily mean that ABCDEF are together - I would post a solution for that later)


kapslock, i mentioned it, because in you statement above you said that ALL couples sit together doesnt mean that ALL sit together, but that AB and/or and/or CD and/or EF sit together. if you stick to this, your calc would be wrong.


Christoph,

You got it right. That's precisely what I meant (and that's what I assumed the question meant when it said all couples sit together, it meant that all the 3 couples were together as AB, CD and EF and not necessarily as ABCDEF).

Let me explain how I got my answer.

When a couple (say A and B) is together, it can be considered one group. Thus, AB becomes a group. This reduces the total number of "groups" whose permutations we're considering to 8 (AB, C, D, E, F, G, H, I). Of course since AB and BA are considered different permutations, total permutations = 8! x 2.

When we consider another couple (say CD) being a group, the number of "groups" become 7. AB, CD, E, F, G, H, I. This gives us 7! permutations, but since AB and BA are different and so are CD and DC, we have 2 factors of 2 included to make it 7!x2x2.

Extend this to three couples sitting together, and you get an answer of 6!x2x2x2.

I still used almost the same reasoning to reach the same answer. I'd appreciate you letting me know the flaw. It'd help me improve the reasoning.

Thanks.
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 [#permalink] New post 13 Apr 2005, 01:50
Then it was correct!! 8-) 8-)
This is difficult to explain
In a few words,
8!*2 are total possible ways in which at least couple sits together (A)
7!*2*2 are total possible ways in which at least 2 couples sit together (B)
6!*2*2*2 are total possible ways in which 3 couples sit together (C)
So group A contains B and C
group B contains C
what is asked is to find A-B-C
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 [#permalink] New post 13 Apr 2005, 01:55
christoph wrote:
Quote:
(Btw this means the couples sit together, not ALL the couples sit together. For example, if A, B are couples, C, D are couples, E, F are couples, and G H I are happily single, then AB are together, CD are together, and EF are together but does not necessarily mean that ABCDEF are together - I would post a solution for that later)


kapslock, i mentioned it, because in you statement above you said that ALL couples sit together doesnt mean that ALL sit together, but that AB and/or and/or CD and/or EF sit together. if you stick to this, your calc would be wrong.


And yes, for ALL couples sit together meaning ABCDEF being together, then the permutations would be 6!x4!. (This assumes that once grouped together, there's no distinction, and the couples can be in any order within the group).

Like if S = Single and C = Couple

1...2...3...4...5...6...7...8...9
S...C..C..C...C..C...C...S...S

is one such allowed permutation, where the C..C..C..C..C..C might not have AB, CD or DF together, as long as all of them are together.

If you consider that ALL couples are be together means couples should be with each other AND in a group as well, like

AB DC FE G I H

then the number of permutations are 4! x 2 x 2 x 2 x 3!

Please critique these solutions.
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 [#permalink] New post 15 Apr 2005, 00:42
what is OA ? is it as it is discussed above ?

a. None of the Couples are sitting together

9! - 8!*2 (at least 1 couple sits together)

b. Only one couple is sitting together

8!*2-7!*2*2-6!*2*2*2

c. All the couples are sitting together

6!*2*2*2

kaplsock, the arch, i think a. and b. is different, because a. means that not a single couple can sit together and b. means that exactly all three couples should sit together, so a. is not the inverse of b..
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 [#permalink] New post 15 Apr 2005, 12:45
9 people, including 3 couples, are to be seated in a row of 9 chairs.


What is the probability that


a. None of the Couples are sitting together


b. Only one couple is sitting together


c. All the couples are sitting together



A = Couple 1 sitting together
B = Couple 2 sitting together
C = Couple 3 sitting together

1. a = Total - AUBUC

2. b = 3 x (only couple 1 sitting together)
= 3 x (A - A&B - A&C + A&B&C)

3. c = A&B&C
---------------------------------------------------------
Total = 9!
A = B = C = 8! x 2
A&B = B&C = A&C = 6! x 2 x 2
A&B&C = 4! x 2 x 2 x 2

AUBUC = A + B + C - A&B - B&C - C&A + A&B&C
= 3A - 3(A&B) + A&B&C


From here, its easy to plug in the numbers and get the answers.
And ofcourse, divide everything by the Total(9!) in the end, to get the probabilities.
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 [#permalink] New post 15 Apr 2005, 12:49
Sorry, a small error in the previous post. Its corrected below.

A = Couple 1 sitting together
B = Couple 2 sitting together
C = Couple 3 sitting together

1. a = Total - AUBUC

2. b = 3 x (only couple 1 sitting together)
= 3 x (A - A&B - A&C + A&B&C)

3. c = A&B&C
---------------------------------------------------------
Total = 9!
A = B = C = 8! x 2
A&B = B&C = A&C = 7! x 2 x 2
A&B&C = 6! x 2 x 2 x 2

AUBUC = A + B + C - A&B - B&C - C&A + A&B&C
= 3A - 3(A&B) + A&B&C


From here, its easy to plug in the numbers and get the answers.
And ofcourse, divide everything by the Total(9!) in the end, to get the probabilities.
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  [#permalink] 15 Apr 2005, 12:49
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