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Couples on Chairs

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by sanju09 » Sat Feb 21, 2009 2:33 am
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?

A. 1/5
B. 1/4
C. 3/8
D. 2/5
E. 1/2
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by sureshbala » Sat Feb 21, 2009 3:01 am
Total number of ways of seating them is 5! = 120.

Total number of ways in which only one couple (say AB) is together and the other couple (say CD) is not together is 2 x 2 x 3! = 24.
(imagine AB as one unit and keeping CD aside we can arrange AB and the other person (say X) in 2 ways and AB can be altered in 2 ways. After this for the couple CD there are 3 places and they can be seated in 3! ways)

Similarly the total number of ways in which couple CD is together and AB is not together is also 24.

Finally, the total number of ways in which both AB and CD will be together is 3 x 2 x 2 = 12.

So out of the 120 encouragements, 60 are not favorable to us.

Hence the probability = 1/2
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by sanju09 » Sat Feb 21, 2009 3:27 am
sureshbala wrote:Total number of ways of seating them is 5! = 120.

Total number of ways in which only one couple (say AB) is together and the other couple (say CD) is not together is 2 x 2 x 3! = 24.
(imagine AB as one unit and keeping CD aside we can arrange AB and the other person (say X) in 2 ways and AB can be altered in 2 ways. After this for the couple CD there are 3 places and they can be seated in 3! ways)

Similarly the total number of ways in which couple CD is together and AB is not together is also 24.

Finally, the total number of ways in which both AB and CD will be together is 3 x 2 x 2 = 12.

So out of the 120 encouragements, 60 are not favorable to us.

Hence the probability = 1/2
Why not D, Suresh?
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by billzhao » Mon Feb 23, 2009 7:34 am
Answer is (D)

The total number of arrangements is: P5=120

(1.) Both couples are sitting adjacent to each other: P3*P2*P2=3!*2*2=24
(2.)1 couple is sitting adjacent to each other: P4*P2=4!*2=48
(3.) The other couple is sitting adjacent to each other: P4*P2=48
(4.) At least one couple is sitting adjacent to each other: (2.)+(3.)-(1.)=48+48-24=72

So the number of arrangements where no couples are sitting adjacent to each other is 120-72=48

So the probability should be 48/120=2/5, Answer is (D)
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by x2suresh » Mon Feb 23, 2009 10:00 am
billzhao wrote:Answer is (D)

The total number of arrangements is: P5=120

(1.) Both couples are sitting adjacent to each other: P3*P2*P2=3!*2*2=24
(2.)1 couple is sitting adjacent to each other: P4*P2=4!*2=48
(3.) The other couple is sitting adjacent to each other: P4*P2=48
(4.) At least one couple is sitting adjacent to each other: (2.)+(3.)-(1.)=48+48-24=72

So the number of arrangements where no couples are sitting adjacent to each other is 120-72=48

So the probability should be 48/120=2/5, Answer is (D)
agreed.

Good approach.
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