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# Probability

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Senior Manager
Joined: 01 May 2004
Posts: 337
Location: USA
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Kudos [?]: 22 [0], given: 0

Probability [#permalink]  14 May 2004, 17:20
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
Manager
Joined: 07 May 2004
Posts: 183
Location: Ukraine, Russia(part-time)
Followers: 2

Kudos [?]: 9 [0], given: 0

Re: Probability [#permalink]  15 May 2004, 00:02
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.
Senior Manager
Joined: 01 May 2004
Posts: 337
Location: USA
Followers: 1

Kudos [?]: 22 [0], given: 0

[#permalink]  15 May 2004, 10:03
Good job!
It's all correct.
Manager
Joined: 07 May 2004
Posts: 183
Location: Ukraine, Russia(part-time)
Followers: 2

Kudos [?]: 9 [0], given: 0

Hi! [#permalink]  15 May 2004, 10:26
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, 10:26
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# Probability

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