Seven people , two of them couples , are seated randomly in

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Seven people , two of them couples , are seated randomly in [#permalink]

08 Oct 2003, 03:28

Seven people , two of them couples , are seated randomly in a row of 7 chairs. Whats the probability that the none of the couples are seated together?

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Re: Probability...Couples [#permalink]

08 Oct 2003, 13:54

praetorian123 wrote:

Seven people , two of them couples , are seated randomly in a row of 7 chairs. Whats the probability that the none of the couples are seated together?

Total ways to seat 7 people = 7! = 5040
Ways that the couple are not together = 7! - (6!*2!) =
5040 - 1440 = 3600

couple not together/total = 3600/5040 = 360/504 = 5/7

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Re: Probability...Couples [#permalink]

08 Oct 2003, 14:29

Quote:

Total ways to seat 7 people = 7! = 5040
Ways that the couple are not together = 7! - (6!*2!) =
5040 - 1440 = 3600
couple not together/total = 3600/5040 = 360/504 = 5/7

Ok, i get a different answer..but i am not real good a probability either..so please check my work
All couples together => total permutations 5! ( 2 couples and 3 kids)
one couple can be seated together in => 2 ! ways
second couple can be seated together in => 2 ! ways
Both the couples can be seated in together => 2! ways

So total = 5! * 2! * 2! * 2! = 6!

i get => 7! - 6! / 7! = 5040-720 / 5040 = 4320 / 5040 = 432 /504

thanks
praetorian

Last edited by Praetorian on 08 Oct 2003, 16:49, edited 1 time in total.

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Re: Probability...Couples [#permalink]

08 Oct 2003, 16:27

exy18 wrote:

praetorian123 wrote:

Seven people , two of them couples , are seated randomly in a row of 7 chairs. Whats the probability that the none of the couples are seated together?

Total ways to seat 7 people = 7! = 5040
Ways that the couple are not together = 7! - (6!*2!) =
5040 - 1440 = 3600

couple not together/total = 3600/5040 = 360/504 = 5/7

I solved the above as if there was only 1 couple.

With 2 couples - the ways that they are not together = 7! - (6!*2!*2!)
=5040 - 2880 = 2160

2160/5040 = 216/504 ??? not sure if this is correct either ??

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[#permalink]

08 Jan 2004, 09:05

7 people with two couple means 4+3 people
If they all the couple are seated together then
C1 C2 S1 S2 S3
combinations are 2*2*5!
Probability that none are seated together = 1 - 2*2*5!/7! = 38/42

[#permalink] 08 Jan 2004, 09:05

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Seven people , two of them couples , are seated randomly in

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Seven people , two of them couples , are seated randomly in

Seven people , two of them couples , are seated randomly in

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