Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated
[#permalink]
Show Tags
25 Nov 2010, 21:55
4
5
shrive555 wrote:
In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?
24 48 56 72 96
When the constraint on an arrangement says, "Two people should not be seated together," we do just the opposite. We make them sit together! Kind of tie them with a thread and assume they are one unit! Let's see why....
These 5 people can be arranged in 5! ways. These are the total number of ways you get. Now, when we tie 2 people together, we have only 4 entities to arrange. We can do this in 4! ways. But in each of these entities, the two people can sit in two different ways (AB and BA). So number of ways in which these two people sit together is 4!*2!.
Now, the ways in which these two people will not be together will be 5!- 4!*2! = 4!(5 - 2) = 72
_________________
Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated
[#permalink]
Show Tags
27 Nov 2010, 14:13
craky wrote:
my solution:
3! * 6*2 = 72
3! for Chuck, Don and Ed
6 positions for Ann and Bob: AC AD AE BC BE CE * 2 (as they can switch they positions)
It is not as concise as Karishma's but it works.
Yes, your logic is absolutely fine. If the number of people is manageable, we can directly find the number of ways in which the two of them should not be seated together. You arranged C, D and E in 3! ways.
Attachment:
Ques2.jpg [ 3.56 KiB | Viewed 5944 times ]
The 4 dots show 4 positions for Ann.When Ann occupies one of these positions, 3 positions are left over for Bob. So Ann and Bob can sit in 4*3 = 12 ways. Total 3! * 12 = 72 ways.
_________________
Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated
[#permalink]
Show Tags
13 Oct 2015, 20:24
1
tuanquang269 wrote:
In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?
(A) 24
(B) 58
(C) 56
(D) 72
(E) 96
In these types of questions, always try to subtract the not allowed arrangements from the total arrangements. This makes our lives much easier. Here we will assume that Ann and Bob always sit together i.e. they are a single entity Total arrangements: 5!
Arrangements in which Ann and Bob sit together = 4!*2! We get this because, total arrangements =4! and Ann and Bob can be arranged in 2! ways
Subtracting this from the total: 5! - 4!*2! = 72. Option D
Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated
[#permalink]
Show Tags
19 Apr 2017, 16:02
1
1
shrive555 wrote:
In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?
A. 24 B. 48 C. 56 D. 72 E. 96
We can use the following formula:
Total number of ways to arrange the 5 people = (number of arrangements when Bob sits next to Ann) + (number of arrangements when Bob does not sit next to Ann)
Let’s determine the number of arrangements when Bob sits next to Ann.
We can denote Ann, Bob, Chuck, Don, and Ed as A, B, C, D, and E, respectively.
If Ann and Bob must sit together, we can consider them as one person [AB]. For example, one seating arrangement could be [AB][C][D][E]. Thus, the number of ways to arrange four people in a row is 4! = 24.
However, we must also account for the ways we can arrange Ann and Bob, that is, either [AB] or [BA]. Thus, there are 2! = 2 ways to arrange Ann and Bob.
Therefore, the total number of seating arrangements is 24 x 2 = 48 if Ann and Bob DO sit next to each other.
Since there are 5 people being arranged, the total number of possible arrangements is 5! = 120.
Thus, the number of arrangements when Bob does NOT sit next to Ann is 120 - 48 = 72.
Answer: D
_________________
Jeffery Miller Head of GMAT Instruction
GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions
Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated
[#permalink]
Show Tags
21 Jul 2018, 09:21
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
close
62% (00:58) correct 
