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

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Intern
Joined: 07 Aug 2003
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Two couples and one single person are seated at random in a [#permalink]

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12 Dec 2003, 01:06
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?
Director
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12 Dec 2003, 03:16
total number of orderings is 5!=120. Couples can sit together in 3!x2x2=24 ways. The prob they are NOT together is 1-24/120=4/5..
Director
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12 Dec 2003, 05:37
agree that there are 120 possible , but are you sure that there are only 24 ways for either couple to be seated together?

If the first couple is in the first two seats, there are two ways for them to be arranged, HW or WH, and there are six ways for the others to be seated. That's 12.
repeat for seats 2&3, 3&4, and 4&5. You get 48 ways in which the first couple can be together.

Repeat for couple #2, and you get a total of 96 ways for them to be together, minus all the ways that both couples are together (double counting).

P, c1, c2 (*2*2)
P, c2, c1 (*2*2)
c1, p, c2 (*2*2)
c1, c2, p (*2*2)
c2, p, c1 (*2*2)
c1, c1, p (*2*2)

96-48 =48.

prob no couple is together= (120-48)/120 60%
CEO
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13 Dec 2003, 00:51
preyshi 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?

total # of ways of seating two couples and one single person = 5! =120

total # of ways both couples can sit together = 3!*2!*2!= 24 ways

total # of ways ONLY one couple can sit = 4! * 2! - 3! *2*2 = 24

(Subtract the # of ways where we have both couples sitting together)

favorable ways = 120- 24 -24 = 72 ways

total ways = 120

probability = 72/120 = 3/5
Director
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13 Dec 2003, 02:21
Ooops!..I missed the ways when there is one couple sitting together..thanx guys..
13 Dec 2003, 02:21
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