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Re: Probability: 9 People, 3 couples and 9 chairs [#permalink]

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11 Apr 2005, 13:10

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 _________________

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).

Re: Probability: 9 People, 3 couples and 9 chairs [#permalink]

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12 Apr 2005, 03:22

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... _________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Re: Probability: 9 People, 3 couples and 9 chairs [#permalink]

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12 Apr 2005, 16:35

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 _________________

you know, I learnt the existence of this kind of math here , never studied it... now I'll give it a try

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

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 !!!

(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. _________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

(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.

Then it was correct!!
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

(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).

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.. _________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

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. _________________

Anyone who has never made a mistake has never tried anything new. -Albert Einstein.

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. _________________

Anyone who has never made a mistake has never tried anything new. -Albert Einstein.