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# Two couples and one single person are seated at random in a

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CEO
Joined: 15 Aug 2003
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Two couples and one single person are seated at random in a [#permalink]  23 Sep 2003, 13:53
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Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?
Intern
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[#permalink]  24 Sep 2003, 06:16
Number of ways 5 people can be seated = 5!
Number of ways in which both the couples sit together = 3!*2*2
Number of ways in which one couple sit together = 4!*2 -3!*2*2

Hence the probability is 3/5.

Is my approach correct ?
Intern
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[#permalink]  25 Sep 2003, 04:48
Is it 1/10?

This is not very elegant, but here goes:
Ways to seat one or both couples together
11223
22113
11232
22131
11322
22311

Multiply by 2 because you can have the same combinations backward (due to symmetry). 6x2=12 ways to seat one or both couples together.

Total number of ways = 5!
12/5! = 1/10
Manager
Joined: 10 Jun 2003
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[#permalink]  25 Sep 2003, 09:43
First 5!/(2!2!) = 30 combinations

10 of these satisfy NEITHER of the couples being adjacent.

10/30 = 1/3
_________________

Sept 3rd

Manager
Joined: 02 Jul 2003
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[#permalink]  25 Sep 2003, 11:09
This is a problem from P.R. that I always miss the answer is

8+8+8+8+16/5!

You take the single person and put him in the first seat and see how many people can sit in the 2nd-5th seat. If the single person starts in the first seat than any of the 4 remaining can sit in the second. Only one person can sit in the third (The person who would not form a couple with the person in the second seat.) and 2 ways for the fourth leaving one way for the fifth seat. thus 1x4x2x1+8. You have to do this process four more times.(Put the single person in seats 2-5) If anyone has an easier solution let me know. this way takes 3-4 minutes which is too long.

Rich
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[#permalink]  25 Sep 2003, 12:16
Okay, I'll try again. There are some glaring faults with my previous answer... I guess it's a good thing I scheduled an afternoon test instead of a morning one!

combinations of one or both couples together:
11322
11223
22113
11232
22131
12231
21132

multiply by 2 because of symmetry(e.g. 11322 flipped around is 22311): 7x2= 14.
multiply by two because couple one is nondistinct.
multiply by two because couple two is nondistinct.
14x2x2=56
Total ways to arrange couples = 5!=120

56/120=7/15
Probability of neither couple sitting together= 1-(7/15) = 8/15

I feel like I'm still missing something though...

praetorian123 wrote:
jlyngal wrote:
Is it 1/10?

This is not very elegant, but here goes:
Ways to seat one or both couples together
11223
22113
11232
22131
11322
22311

Multiply by 2 because you can have the same combinations backward (due to symmetry). 6x2=12 ways to seat one or both couples together.

Total number of ways = 5!
12/5! = 1/10

nope...not correct.
Intern
Joined: 23 Aug 2003
Posts: 19
Location: ny
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[#permalink]  25 Sep 2003, 14:17
Okay, I think this will be my last attempt...

final answer = 2/5

Combinations of one or both couples sitting together:
11223
22113
11322
12213
21123
12231
21132
22131
11231

Multiply by 2 for symmetry: 9x2=18
Multiply by 2 because couple 1 is nondistinct: 18x2=36
Multiply by 2 because couple 2 is nondistinct: 36x2=72

Total ways to arrange people: 5! = 120
72/120 = 3/5

1-(3/5) = 2/5

I think my approach is legitimate... it's just so hard to think of different combinations.

I just tried rich's method, and it seems like that's the best one:
Arranging so couples don't sit together...
single person in seat 1: 1x4x1x2x1=8
single person in seat 2: 4x1x2x1x1=8
single person in seat 3: 4x2x1x2x1=16
single person in seat 4: 4x2x1x1x1=8
single person in seat 5: 4x2x1x1x1=8
8+8+16+8+8 = 48
48/5!=2/5

As a shortcut, you could see that the number of ways to arrange people when the single person is in seat 1 is the same as the the number of ways when the single person is in seat 5. The same is true when the single person is in seat 2 or seat 4.

That method actually doesn't take very long... I'd say about a minute.
Intern
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[#permalink]  29 Sep 2003, 01:45
I think it could be 5 ways.

Whats the correct answer???
Senior Manager
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[#permalink]  29 Sep 2003, 09:43
This is what I would do:

Prob of neither couples sitting together= 1- Prob of both couples sitting together

Prob of both couples sitting together:

favorable outcomes=6
Possible outcomes=5!/2!*2!=30

Solution: 1-(6/30)=24/30=0.8

How close was I?
CEO
Joined: 15 Aug 2003
Posts: 3470
Followers: 61

Kudos [?]: 697 [0], given: 781

[#permalink]  29 Sep 2003, 10:43
MartinMag wrote:

This is what I would do:
Prob of neither couples sitting together= 1- Prob of both couples sitting together
Prob of both couples sitting together:
favorable outcomes=6
Possible outcomes=5!/2!*2!=30

Solution: 1-(6/30)=24/30=0.8
How close was I?

Oh..close..but not quite. The Answer is 2/5.

Thanks
Praetorian
[#permalink] 29 Sep 2003, 10:43
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# Two couples and one single person are seated at random in a

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