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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 02 Apr 2006, 16:30
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?

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 [#permalink] New post 02 Apr 2006, 16:37
yes I was wrong earlier.

its 3/5

P(A) = Lets assume that all couple needs to be seated together always.

Hence single person can either seat at first place or last place.

Case 1.
When single person seats at the first place.
Number of ways in which 2 couple can seat at remaining places = 4 * 3 *2 *1.

case 2:
When single person seats at the last place
Number of ways in which 2 couple can seat at remaining places = 4 * 3 *2 *1.

Combining case 1 and case 2.

Total number of ways 2 couples can seat together = 2 * 4* 3*2 *1

Total number of ways 5 people can seat = 5* 4*3 *2 *1

P(A) = 2/5

Hence probability that 2 couple will never seat together = 3/5

Last edited by gmat_crack on 02 Apr 2006, 18:24, edited 1 time in total.
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 [#permalink] New post 02 Apr 2006, 16:48
Nope,

give an explanation please
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Re: Probability [#permalink] New post 02 Apr 2006, 17:08
jodeci 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 sits together in adjacent chairs?

Thanks


I got 9/10
let's assume we have 5 people numbered 1 to 5 and we want to keep

(1,2) and (4,5) together therefore 3 must be first, in the middle or last

and we can order (1,2) in two ways
(4,5) in two ways

so it would be (2*2*3)/5!

equal to 1/10

so 1 - 1/10

is 9/10
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 [#permalink] New post 02 Apr 2006, 17:52
i got 4/15

the single person has to sit either first middle or last

case 1

first

1(because there is only 1 single person )*4(because any of the other 4 people can sit next to him ) * 2 (because out of the three only 2 that are not coupled to 4 can sit next to 4) *1( because only the person who isnt coupled to 2 can sit next to 2) and times 1 because only one person is left ) = 8

case 2
last

4*2*1*1*1 =8

case 3
and middle

4*2*1*2*1= 16

now total possible arrangements is 5! = 120

(16+8+8)/120 = 32/120 and this is = to 4/15

hope i didnt screw it up ! :roll:

and 9/10 just seems a bit too close to 1 to be correct
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 [#permalink] New post 02 Apr 2006, 18:02
i think i know what i did wrong the single person can realy sit anywhere ...

lets say we have these people

A1 A2 B1 B2 S

it could be

S A1 B1 A2 B2

or

A1 S B1 A2 B2

and so on ... lol now i should figure our how 2 write this down
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 [#permalink] New post 02 Apr 2006, 18:42
2/5.

Both sit together = 2*2*3! = 24.
One sit together = 2*4*3*2*1 = 48 => Since there are two couple..total nubmer of ways - 48*2 = 96.

Number of ways that at least one couple sites together = 96-24 = 72.

Number of ways that none of couple sits together = 5! - 72 = 48.

Prob. = 48/120 = 2/5.
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 [#permalink] New post 02 Apr 2006, 18:45
The case of any couple sitting togthere = 4!*2

Since there 2 couples it is 2*4!*2

But you need to discard the case where the 2 couples are sitting together
(since we have counted that twice in the above case)

So

total -(2*Single couple sitting together) + 2 couples sitting together

120 -(2*4!*2) + 3!*2!*2!
=> 48

So probability of sitting together 48/120

So neither case = 72/120 = 3/5
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 [#permalink] New post 02 Apr 2006, 19:21
OA is 2/5 or 48/120

lhotseface

Please explain your answer further..
Thanks
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 [#permalink] New post 02 Apr 2006, 19:33
oops it is 48/120....

OK the solution is like this

Let (A,B) , (C,D) and E be the people . People in () are couples

Total # of arrabgemnets = 120

Lets take the case of (AB) together So the arrgments are 4!*2!(since AB can be BA)

Same case ith (CD)

So now the case where only a single couple(either AB or CD) is together = 2*4!*2! ---------------(1)

And the case where the 2 couples are together = 3!*2!*2!-----(2)


But (2) has been counted twice in (1)
So we need to add back (2)

Finally it is 5! - (2*4!*2!) + (3!*2!*2!) = 48

Probabilty = 48/120 = 2/5
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 [#permalink] New post 02 Apr 2006, 20:03
lhotseface wrote:
2/5.

Both sit together = 2*2*3! = 24.
One sit together = 2*4*3*2*1 = 48 => Since there are two couple..total nubmer of ways - 48*2 = 96.

Number of ways that at least one couple sites together = 96-24 = 72.

Number of ways that none of couple sits together = 5! - 72 = 48.

Prob. = 48/120 = 2/5.


Hi nice explanation. Can you point out where I'm making mistake.
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 [#permalink] New post 02 Apr 2006, 20:47
At first glance, it looks like you missed the scenario that the single person can sit in the middle even when the couples sit together...

gmat_crack wrote:
lhotseface wrote:
2/5.

Both sit together = 2*2*3! = 24.
One sit together = 2*4*3*2*1 = 48 => Since there are two couple..total nubmer of ways - 48*2 = 96.

Number of ways that at least one couple sites together = 96-24 = 72.

Number of ways that none of couple sits together = 5! - 72 = 48.

Prob. = 48/120 = 2/5.


Hi nice explanation. Can you point out where I'm making mistake.
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 [#permalink] New post 03 Apr 2006, 06:54
lhotseface wrote:
At first glance, it looks like you missed the scenario that the single person can sit in the middle even when the couples sit together...

gmat_crack wrote:
lhotseface wrote:
2/5.

Both sit together = 2*2*3! = 24.
One sit together = 2*4*3*2*1 = 48 => Since there are two couple..total nubmer of ways - 48*2 = 96.

Number of ways that at least one couple sites together = 96-24 = 72.

Number of ways that none of couple sits together = 5! - 72 = 48.

Prob. = 48/120 = 2/5.


Hi nice explanation. Can you point out where I'm making mistake.


Thanks buddy. Got it. :wink:
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 [#permalink] New post 04 Apr 2006, 11:13
I think this method is also applicable.

Say that A.B, C.D are couples and E is single

Ways that they are adjacent to each other

A.B, C.D, E = 4*3*1*2*1 = 24

A.B, E ,C.D =4*3*1*2*1 = 24

E, A.B, C.D =1*4*3*2*1 = 24

Add them up= 24*3 = 72

Total ways to arrange 5 people in a row = 5!

5!-72= 48

Probability = 48/120 = 2/5
  [#permalink] 04 Apr 2006, 11:13
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