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:
Christina scored 760 by having clear (ability) milestones and a trackable plan to achieve the same. Attend this webinar to learn how to build trackable milestones that leverage your strengths to help you get to your target GMAT score.
Join a free live webinar and learn the winning strategy for a 700+ score on GMAT & the perfect application. Save your spot today! Wednesday, March 27th at 3 pm PST
Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated
[#permalink]
Show Tags
25 Nov 2010, 21:55
4
6
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
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
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 7573 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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
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.
Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated
[#permalink]
Show Tags
23 Nov 2018, 21:08
This is a classic FCP with restrictions problem. It's faster to use the "complement" to solve these problems than to count the arrangements directly. If we can count the number of total arrangements, then subtract the number of cases we don't want, we're left with the number of cases we do want.
[# of arrangements with Ann & Bob not sitting together]
= [# without restrictions] - [# with Ann & Bob sitting together]
When arranging 5 people where order matters, the # of arrangements is 5! = 120.
To calculate the # of arrangements where Ann & Bob are seated together, we "glue" them together as a single unit: [Ann, Bob]. In that case we're arranging 4 units where order matters, that's 4! = 24.
But wait! Ann, Bob, Chuck, Don, Ed is different from Bob, Ann, Chuck, Don, Ed even though Ann and Bob are still seated together before the rest. So, there's an additional 24 arrangements where we have them glued together as [Bob, Ann].
The # of arrangements with Ann & Bob not sitting together is 120 - 24 - 24 = 72.
For a review, on this "complement" concept check out the FCP with Restrictions lesson video.
_________________
Thanks! Do give some kudos.
Simple strategy: “Once you’ve eliminated the impossible, whatever remains, however improbable, must be the truth.”
close
61% (01:23) correct 
