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

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Intern
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9 people, including 3 couples , are to be seated in a row of [#permalink] New post 08 Dec 2003, 23:27
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 together
b. Only one couple is together
c. All the couples are together
Intern
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my take on it [#permalink] New post 09 Dec 2003, 00:40
Hi, thanks for the nice question

1. P(couples not together) = 1 - P(couples together)
1-((2*2*2*6!)/9!) = 1 - 8/(9*8*7) = 62/63

2. 1 - P(2 couples together)
= 1 - (3*2!*2!*7!)/9! = 15/18

3. 1/63

Are these correct ?
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 [#permalink] New post 09 Dec 2003, 03:40
For part 2 I used a different approach total orderings 9!, one couple can be selected in 3 ways, can sit in 8 ways and occupy the 2 places in 2 ways, as the rest of the people can sit in 7! ways OR 3x8x2x7!/9!=6/9=2/3 it is a little bit bigger than 15/18 but guess this is OK since this is probability after all... :)
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Re: Probability [#permalink] New post 13 Dec 2003, 01:19
preyshi 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 together
b. Only one couple is together
c. All the couples are together


Quote:
a. None of the Couples are together


(1)total # of ways = 9!

(2)One couple is together => 8! *2 - 7!*2*2 - 6!*2*2*2

(3) two couples are together => 7!*2*2 - 6! *2*2*2 ways

(4)three couples are together => 6! *2*2*2 ways

none are together = (1) - (2) - (3) - (4) / (1)


Quote:
b. Only one couple is together


(2) One couple is together => 8! *2 - 7!*2*2 - 6!*2*2*2

prob => (2) / 9!

Quote:
c. All the couples are together


(4) three couples are together => 6! *2*2*2 ways

prob => (4) /9!
Re: Probability   [#permalink] 13 Dec 2003, 01:19
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