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Probability [#permalink] New post 14 May 2004, 18:20
Hello!
Guys, try to solve this problem

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
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Re: Probability [#permalink] New post 15 May 2004, 01:02
boksana wrote:
Hello!
Guys, try to solve this problem

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


Hi, boksana!

All combinations: 9!

Combinations for which all the couples are together:

6!*2*2*2 (because there are 6! ways to arrange 6 entities - 3 couples and 3 other persons, and 2 ways of sitting a couple MW and WM).

=> p(c) = 6!*8/9! = 1/63.

Combinations for which 2 couples are together and 1 is not:

7!*2*2 - 6!*2*2*2 (the same logic - 7 entities can be seated in 7! ways, and there are 2 ways in each couple).

=> p(.) = (7!*4-6!*8)/9! = 1/18-1/63=5/126.

Combinations for which only 1 couple is:

8!*2 - [previous 2 cases together] = 8!*2 - 7!*4.

p(b) = [8!*2-7!*4]/9! = 2/9-1/18 = 1/6.

Combinations for which there are no couples:

p(a) = 1 - p(b) - p(.) - p(c) = 1 - 1/6 - 5/126 - 1/63 = 98/126 = 7/9.
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 [#permalink] New post 15 May 2004, 11:03
Good job!
It's all correct.
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Hi! [#permalink] New post 15 May 2004, 11:26
boksana wrote:
Good job!
It's all correct.


I know... :-)

Are there any questions? Proposals? Suggestions?

Feel free to ask if you have any.
Hi!   [#permalink] 15 May 2004, 11:26
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