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Those who intend to take PR I plz do not read this

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Those who intend to take PR I plz do not read this [#permalink] New post 31 Jan 2005, 03:14
Those who intend to take PR I plz do not read this question...
The question is simple and I understood the combinations too. but I am looking forward for formula and explaination?

Two couples and one single are seated in random in a row of five chairs.
What is the probability that neither of the couples sit together in an adjacent chair?

1/5
1/4
3/8
2/5
1/2
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 [#permalink] New post 31 Jan 2005, 07:03
My permutation and combination is not very good, but let me try:

P(5,5)-2*(C2,1)*P(4,4)=5!-4*4!=4!
4!/P(5,5)=4!/5!=1/5

(A)
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 [#permalink] New post 31 Jan 2005, 07:10
Actually, that's not right.

A couple sits together P(2,2)*P(4,4)
But both couple sits togeter P(2,2)*P(2,2)*P(3,3)

So no couple sits together would be
P(5,5)-2P(2,2)*P(4,4)+P(2,2)*P(2,2)*P(3,3)=5!-4*4!-4*3!=5!-3*4!=2*4!
Probability=2*4!/5!=2/5 :?:
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 [#permalink] New post 31 Jan 2005, 07:16
i am getting E..

1/2
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 [#permalink] New post 31 Jan 2005, 07:22
Can you show me how to do this one?
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 [#permalink] New post 31 Jan 2005, 07:26
actually I am wrong it will be 2/5

M1 F1, M2 F2, S

the way its better approached is takign the single person in my case being S

S can be put in 5! ways..
S _ _ _ _
_ S _ _ _
_ _ S _ _
_ _ _ S _
_ _ _ _ S

the others can be filled as follows..

S 4 2 1 1 = 8
4 S 2 1 1 = 8
4 2 S 2 1 = 16
4 2 1 S 1 = 8
4 2 1 1 S = 8

so total will be 48
therefore, 48/120=2/5

Last edited by vprabhala on 31 Jan 2005, 07:32, edited 1 time in total.
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 [#permalink] New post 31 Jan 2005, 07:31
What about M2 and F2? The way you are doing it you can't make sure they don't sit together?
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 [#permalink] New post 31 Jan 2005, 07:32
HongHu, i edited the post...
it is 2/5...
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 [#permalink] New post 31 Jan 2005, 19:52
# of ways both couples can sit together:
2 ways to pick 2 couples
4 ways they can be seated so that the couples are adjacent.
3! ways to seat the 2 couples with the single person
so total = 2.4.3! = 48

# of ways 1 couple can sit together
2 ways to pick 1 couple from 2
2 ways they can be seated adjacent to each other
now this couple need to be seated with 3 other individuals with the constraint that the other couple cannot sit together, which is similar to seating 4C2 seating.
so total = 2.2.4C2 = 24

Total number of ways couple can sit together = 72

Problem is asking for Pb(not sitting together) = 5! - 72/5! = 48/120 = 2/5.
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 [#permalink] New post 01 Feb 2005, 12:53
Yes, The OA is D(2/5).

Thanks for the explaination guys. I got it....
Vijo.
  [#permalink] 01 Feb 2005, 12:53
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