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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 12 Oct 2003, 17:08
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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?

Answer : 2/5

i started from the basic formula

[n(A) + n(B)- n(A and B)+n(not A and not B) ] / (total # of ways) = 1

P(A) + P(B) - P(A and B) + P(not A and Not B) = 1

in our case
n(A) = # of ways one of the two couples seated together
n(B) = # of ways the second of the two couples can beseated together
n(A and B) = # of ways Both of the couples can be seated together
n(Not A and Not B) = # of ways None of the couples together

In all these cases, the total # of ways is 5!, so we divide the entire sum by 5!

A bit of work will give

P(Neither couples seated together ) = 1 -[ P(one couple seated together) + P(second couple together) - P(both couples seated together)]


P(one couple seated together)

We count the couple as a one single unit, as they have to be together.
so we now have a total of 4 "people". they can be arranged in 4! ways. now the couple itself can be arranged in 2! ways. The total number of possible arrangements is 5! ... it follows that

P(one couple seated together) = 4! * 2/ 5! = 48/120

P(second couple seated together) => same as the above = 48/ 120

P(two couples seated together)
so now we have three "people". using the same logic as above,
3 people can be arranged in 3! ways. Note : Here we multiply first by 2 as the first couple can be arranged together in 2! ways...similarly the second couple can be seated together in 2! ways...and the total # of arrangements is 5!.
P(two couples seated together ) = 3! * 2 * 2 / 5 ! = 24 /120

So , it follows that
P(neither couple seated together) = 1 - 48/120 - 48/120 + 24/120 = 1- 72/120 = 48/120 = 2/5

you think there is a faster way or is this fine?

thanks
praetorian
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Re: PS : Probability (Couples) ..check my work please [#permalink] New post 13 Oct 2003, 01:01
stolyar, could you show us a faster method , if there is one?

thanks
praetorian
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 [#permalink] New post 15 Jan 2004, 19:23
Hi Praetorian

Let the couples be C1,C2 and S
We can arrange these in 3! ways
no of ways atleast a couple can sit together = 3! * 2 * 2= 4!

No couple sits together = 1 - 4!/5! = 4/5

What did I do wrong here ?
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 [#permalink] New post 16 Jan 2004, 03:32
anandnk you have calculated only the option when 2 couples are together and they can sit in 3! x2x2=24 ways. But we have to calculate also the option when 1 couple is together , then the possible ways for the couples to sit are 4!x2=48. So when 2 couples are together we have 24 ways and when 1 is together we have 48 or total 72 ways. Drawing venn diagrams is helpful here . the prob is 1-72/120=48/120=2/5
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 [#permalink] New post 16 Jan 2004, 05:40
opps I got it.
But I have a question.
Let us say 1 couple C1 sits together then we have h2(husband2),w2(wife2), s(lonely person) members left
now we have C1, h1,w1,s people sitting in 4! ways and C1 can sit in 2 ways. so we have 2 * 4! ways

In this we also have the combination C1,h1,w1,s and C1,w1,h1,s. These
combinations account for 2 couples sitting together. Why do we again add 24( combinations in which 2 couples sit together) ?
  [#permalink] 16 Jan 2004, 05:40
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