BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

good prob

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 371
Joined: Tue Apr 29, 2008 10:16 am
Thanked: 6 times
Followed by:1 members

good prob

by vaivish » Sun Jun 01, 2008 12:04 pm
6 persons seat themselves at round table. What is the probability that 2 given persons are adjacent?

(A) 1/5

(B) 2/5

(C) 1/10

(D) 1/7

(E) 2/15
Top
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 127
Joined: Sat Apr 19, 2008 1:02 am
Thanked: 12 times

by jasonc » Sun Jun 01, 2008 12:26 pm
unless I'm mistaken (could be since I'm really tired haha) the problem should be pretty straight forward, and the answer should be 2/5

for any given 2 people, regardless of where the first person sits, the second person has 2 possible ways to be next to the first person (L & R of the person) and 5 possible ways of seating himself total = 2/5
I beat the GMAT! 760 (Q49/V44)
Top
Master | Next Rank: 500 Posts
Posts: 167
Joined: Tue Apr 22, 2008 12:48 am
Thanked: 15 times

by durgesh79 » Mon Jun 02, 2008 9:36 pm
i have a doubt in thse type of cirular table questions,

Suppose there are 5 people ABCDE,

If I ask everyone to standup and move to one postion left (or right), will it be a new arrangemnet.

I'll put the question in other words. Are following arrangements different or same in a circular table question.

ABCDE
BCDEA
CDEAB
DEABC
.......
Top
Legendary Member
Posts: 631
Joined: Mon Feb 18, 2008 11:57 pm
Thanked: 29 times
Followed by:3 members

by netigen » Mon Jun 02, 2008 10:56 pm
In a circular table these are the same.

There is a formula to find the different possible permutations in a circular table with n slots

total possibilities = (n-1)!
Top
Legendary Member
Posts: 631
Joined: Mon Feb 18, 2008 11:57 pm
Thanked: 29 times
Followed by:3 members

by netigen » Mon Jun 02, 2008 10:57 pm
Read up on this - https://www.mansw.nsw.edu.au/members/ref ... no4yen.htm
Top
Master | Next Rank: 500 Posts
Posts: 167
Joined: Tue Apr 22, 2008 12:48 am
Thanked: 15 times

by durgesh79 » Tue Jun 03, 2008 12:21 am
Thanks netigen,

Coming back to the question,

6 people can be arranged in 5! ways
if we consider two people always sitting together as a single unit, we have 5 people around the table. Can be arranged in 4! ways.
These 2 person can change their position in 2 ways.

so the answer = 2 * 4! / 5! = 2/5
Top
Post new topic Post Reply
• Page 1 of 1