Princeton Review Exam Question (Probability)

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abbyyip: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Mon May 07, 2007 3:28 pm; Location: Pasadena, CA

Princeton Review Exam Question (Probability)

by abbyyip » Tue Jun 19, 2007 7:14 pm

Does anyone know a simple or clear way to solve this Probability Problem?

Two couples and one single person are seated at random in a row of 5 chairs. What is the probability that niether of the couples sits together in adjacent chairs?

a) 1/5
b) 1/4
c) 3/8
d) 2/5
e) 1/2

Thanks!

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Source: Beat The GMAT — Problem Solving | Tue Jun 19, 2007 7:14 pm

Neo2000: Legendary Member; Posts: 519; Joined: Sat Jan 27, 2007 7:56 am; Location: India; Thanked: 31 times

by Neo2000 » Wed Jun 20, 2007 7:43 am

No: of ways of arranging 5 people in 5chairs = 5! = 120

Say the two couples are A,B and C,D
Now, treating the two couples as 1 unit each, there are 3units of people to be arranged in 3chairs. This can be done in 3! = 6ways
Each unit of two people among themselves can be arranged in 2x2! = 4ways

So total no: of ways = 6x4 = 24

There is also the possibility that one couple is seated together and the other is not. Select one couple = 2C1 = 2
Now there are 4units ( 1 single man, the chosen couple and the remainin 2 say C,D) to be arranged such that the other two people never sit together
4 can be arranged in 4! ways
Total number of ways of arranging 3units of people = 3! = 6

Total number of ways or arranging 4units of people such that C,D are never together = 24-6 = 18
Since there are 2couples 2x18 = 36

Total number of favourable events = 24 +36= 60
Probability = 60/120 = 1/2

Probability that either couple never sits together = 1- 1/2 = 1/2

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cooldude2281: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Mon Jun 11, 2007 10:36 am

by cooldude2281 » Wed Jun 20, 2007 8:08 am

gr8 explanation

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Neo2000: Legendary Member; Posts: 519; Joined: Sat Jan 27, 2007 7:56 am; Location: India; Thanked: 31 times

by Neo2000 » Wed Jun 20, 2007 8:16 am

cooldude2281 wrote:gr8 explanation

Is it the right answer though?

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mschling52: Master | Next Rank: 500 Posts; Posts: 105; Joined: Mon Sep 18, 2006 12:34 pm; Location: OH; Thanked: 7 times; GMAT Score:780

by mschling52 » Wed Jun 20, 2007 9:33 am

I'm going to go with D - 2/5. There are 120 ways to arrange the 5 people. Call them A,B,C,D,E and let A,B and C,D be the couples and E be the single. I looked at this by fixing the position of the single person and then evaluating how many arrangments of the others are possible. For example, if E is in the first seat, then you have the following

(E)---(4 choices)---(2 choices)---(1 choice)---(1 choice)

The second seat can clearly be A,B,C,or D, then the 3rd must be a member of the other couple, the fourth must be the parter of the person in the second seat, and the last must be the partner of the 3rd. So 4*2*1*1 = 8 arrangements. If E is in the second seat, you get

(4 choices)---(E)---(2 choices)---(1 choice)---(1 choice)

Note the 3rd seat cannot be occupied by the partner of the first since this would leave the other couple to occupy 4 and 5. Again this yields 8 arrangements. If E is in the middle seat, you get

(4 choices)---(2 choices)---(E)---(2 choices)---(1 choice)

for 16 arrangements. Since putting E in the 4th and 5th seat will be the same as the 1st and 2nd seats, just in reverse order, we know the possible outcomes without a couple sitting together are 8+8+16+8+8 = 48. Since the total number of arrangements is 120, the probability is 48/120 = 2/5.