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

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Two couples and one single person are seated at random in a [#permalink] New post 19 Dec 2007, 18:05
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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 sit together in adjacent chairs ?

a. 1/5
b. 1/4
c. 3/8
d. 2/5
e. 1/2
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 [#permalink] New post 19 Dec 2007, 20:52
Expert's post
D

The total number: 5P5=120

12323 + mirror symmetry: 4+4=8
13232 + mirror symmetry: 4+4=8
21323 + 31232+ mirror symmetry: 4+4+8=16
23123 + 32132: 4+4=8
23132 + 32123: 4+4=8

N=8+8+16+8+8=48

p=48/120=2/5
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Re: PS couple seating [#permalink] New post 19 Dec 2007, 21:13
young_gun wrote:
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 sit together in adjacent chairs ?

a. 1/5
b. 1/4
c. 3/8
d. 2/5
e. 1/2


AB is the first couple XY is the second couple

S is the sinlge person

Probability that neither sit together =1- probability that they do sit together.

First 5! is the total possiblities

then SABXY 4!. but we must count ABXYS so 4!*2

48/120 = 2/5

Something doesnt seem right with this answer...

shouldnt I be getting 3/5? since 48/120 is the probability that they do sit together?
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Re: PS couple seating [#permalink] New post 19 Dec 2007, 22:58
Expert's post
GMATBLACKBELT wrote:
...then SABXY 4!. but we must count ABXYS so 4!*2...
Something doesnt seem right with this answer...


I've slightly corrected your way:

1. (AB),X,Y,S: 4P4*2P2=2*4!
2. A,B,(XY),S: 4P4*2P2=2*4!
3. (AB),(XY),S: 3P3*2P2*2P2=2*2*3!=4!
4. p=1-(2*4P4*2P2-3P3*2P2*2P2)/5P5=1-3*4!/(5*4!)=2/5
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 [#permalink] New post 19 Dec 2007, 23:08
Is it 2/5? Pls provide OA along with the source

Thanks
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Re: PS couple seating [#permalink] New post 20 Dec 2007, 08:11
walker wrote:
GMATBLACKBELT wrote:
...then SABXY 4!. but we must count ABXYS so 4!*2...
Something doesnt seem right with this answer...


I've slightly corrected your way:

1. (AB),X,Y,S: 4P4*2P2=2*4!
2. A,B,(XY),S: 4P4*2P2=2*4!
3. (AB),(XY),S: 3P3*2P2*2P2=2*2*3!=4!
4. p=1-(2*4P4*2P2-3P3*2P2*2P2)/5P5=1-3*4!/(5*4!)=2/5


can you clarify that?

1 - ( ??? - repeats) / total
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Re: PS couple seating [#permalink] New post 20 Dec 2007, 08:47
Expert's post
bmwhype2 wrote:
can you clarify that?


1. (AB),X,Y,S: N1=4P4*2P2=2*4!
2. A,B,(XY),S: N2=4P4*2P2=2*4!
3. (AB),(XY),S: N3=3P3*2P2*2P2=2*2*3!=4!
4. p=1-(N1+N2-N3)/5P5=1-3*4!/(5*4!)=2/5

we should extract N3 because we've counted (AB),(XY),S twice in N1 and N2.
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 [#permalink] New post 22 Dec 2007, 01:39
walker wrote:
D

The total number: 5P5=120

12323 + mirror symmetry: 4+4=8
13232 + mirror symmetry: 4+4=8
21323 + 31232+ mirror symmetry: 4+4+8=16
23123 + 32132: 4+4=8
23132 + 32123: 4+4=8

N=8+8+16+8+8=48

p=48/120=2/5



Could you please explain what is 123??? What are these numbers signifying? What do you mean by mirror symmetry?
It'd be so nice of you, if could explain this a bit more.
-Thanks
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 [#permalink] New post 22 Dec 2007, 01:46
Expert's post
LM wrote:
Could you please explain what is 123???


1 - a single person
2,2 - persons of first couple
3,3 - persons of second couple

LM wrote:
What do you mean by mirror symmetry?


12323 ==> 32321 - mirror symmetry (reverse order) It's like we look at that through mirror.
or
a,b,c,d,e,f,g ==> g,f,e,d,c,b,a
  [#permalink] 22 Dec 2007, 01:46
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