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
Couples on Chairs
This topic has expert replies
- sanju09
- GMAT Instructor
- Posts: 3650
- Joined: Wed Jan 21, 2009 4:27 am
- Location: India
- Thanked: 267 times
- Followed by:80 members
- GMAT Score:760
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
- sureshbala
- Master | Next Rank: 500 Posts
- Posts: 319
- Joined: Wed Feb 04, 2009 10:32 am
- Location: Delhi
- Thanked: 84 times
- Followed by:9 members
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
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
- sanju09
- GMAT Instructor
- Posts: 3650
- Joined: Wed Jan 21, 2009 4:27 am
- Location: India
- Thanked: 267 times
- Followed by:80 members
- GMAT Score:760
Why not D, Suresh?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
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
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)
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)
Yiliang
-
- Master | Next Rank: 500 Posts
- Posts: 258
- Joined: Thu Aug 07, 2008 5:32 am
- Thanked: 16 times
agreed.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)
Good approach.