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:
Carolyn and Brett - nicely explained what is the typical day of a UCLA student. I am posting below recording of the webinar for those who could't attend this session.
Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
[#permalink]
Show Tags
04 Mar 2015, 03:28
1
6
00:00
A
B
C
D
E
Difficulty:
45% (medium)
Question Stats:
60% (00:52) correct 40% (01:14) wrong based on 178 sessions
HideShow timer Statistics
Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is the probability that Childeric and Ethelred sit next to each other?
Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
[#permalink]
Show Tags
04 Mar 2015, 18:01
5
Bunuel wrote:
Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is the probability that Childeric and Ethelred sit next to each other?
A. 1/30 B. 1/15 C. 1/5 D. 2/5 E. 7/20
Kudos for a correct solution.
For such questions, use counting techniques to solve the problem. The total arrangements (total outcomes since we are looking for a probability) is 5!, which is our denominator. For the numerator (desired scenarios), club C and E together. So we have A, B, D, CE. These can be arranged in 4! ways and CE can also be arranged as EC, multiply 4! by 2. Therefore, we get 4!*2/5*4! = 2/5, which is answer D.
_________________
"Hardwork is the easiest way to success." - Aviram
One more shot at the GMAT...aiming for a more balanced score.
Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
[#permalink]
Show Tags
04 Mar 2015, 03:52
4
1 2 3 4 5 _ _ _ _ _
Total no of empty space is 5 so total no of ways A B C D and E can sit is 5 !
Solution 1: Now condition is C&E will sit together If they sit at 1st and 2nd place . # of ways will be 1*3! but C&E can be either at 1 or at 2 so # of ways will be 1*3!*2! Similarly they can sit together at position 2 and 3 or 3 and 4 or 4 and 5
So total # ways will be 1*3!*2! + 1*3!*2! + 1*3!*2! + 1*3!*2! = 4 *1*3! *2 !
So probability is (4! * 2! ) / 5! = 2/5
Solution 2: Consider C*E as 1 person . So we have 4 ppl and 4 places . # of ways they can sit is 4! C&E can sit as C&E or E&C so total # of ways is 4! * 2!
Status: It always seems impossible until it's done!!
Joined: 29 Aug 2012
Posts: 1123
Location: India
WE: General Management (Aerospace and Defense)
Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
[#permalink]
Show Tags
04 Mar 2015, 03:56
4
Bunuel wrote:
Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is the probability that Childeric and Ethelred sit next to each other?
A. 1/30 B. 1/15 C. 1/5 D. 2/5 E. 7/20
Kudos for a correct solution.
Ah pretty tricky question.
So the total arrangement will be 5! =120
We can make Childeric and Ethelred sit next to each other in [1st and 2nd] or [2nd and 3rd] or [3rd and 4th] or [4th and 5th]
that's 4 and we can make Ethelred sit first and then Childeric. so total will be 4*2= 8.
Now we can make Anaxagoras, Beatrice and Desdemona sit in other 3 open places in 3! ways= 6
So total will be 8*6= 48
The probability will be \(\frac{48}{120}\)= \(\frac{12}{30}\)= \(\frac{4}{10}\) = \(\frac{2}{5}\)
Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
[#permalink]
Show Tags
08 Mar 2015, 13:22
Bunuel wrote:
Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is the probability that Childeric and Ethelred sit next to each other?
First, we will count all the possible arrangements of the five children on the five seats, all the possible orders. This is 5! = 120. That’s the denominator.
Now, the more challenging part: we have to figure out how many arrangements there are involving C & E sitting together. This is a tricky problem to frame, so I’ll demonstrate the steps to follow. First, let’s look at the seats these two could be next to each other. There are four possible pairs of seats in which they could be next to each other i. X X _ _ _ ii. _ X X _ _ iii. _ _ X X _ iv. _ _ _ X X
In each of those four cases, we could have either CE or EC, either order, so that’s 4 x 2 = 8 ways we could have just C & E sitting next to each other with the remaining three seats empty.
For the final step, we need to consider the other three children, A & B & D. In each of the eight cases, there are three blank seats waiting for those three, and those three could be put in any order in those blank seats. Three elements in any order —- that’s 3! = 6. Thus, the total number of arrangements in which C & E would be next to each other would be 8 x 6 = 48. This is our numerator.
The probability would be this number, 48, over the total number of arrangements of the children, 120.
Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
[#permalink]
Show Tags
29 Aug 2016, 23:03
Total number of arrangements for 5 children to sit in 5 chairs = 5!
Now, consider kids C&E as single entity. Since we consider 2 children as a single entity, we are left with 4 chairs, as the adjacent chairs in any case will be occupied by the two kids, whom we considered as a single entity. Hence we are left with 4 chairs and 4 children = 4!
But since it is unordered we can place C&E or E&C within these two chairs. Hence 2*4!
Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
[#permalink]
Show Tags
08 Nov 2017, 16:42
Bunuel wrote:
Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is the probability that Childeric and Ethelred sit next to each other?
A. 1/30 B. 1/15 C. 1/5 D. 2/5 E. 7/20
We need to determine:
P(Childeric and Ethelred sit next to each other)
This probability is given by (Childeric and Ethelred sit next to each other)/(total number of ways to arrange the 5 children).
The total number of ways to arrange the 5 children is 5! = 120 ways.
The number of ways with Childeric next to Ethelred can be shown as:
[C-E] - A - B - D
We are treating C and E as a single entity because they must sit next to each other. Thus, we have 4 positions that can be arranged in 4! = 24 ways and we also see that [C-E] can be arranged in 2! = 2 ways, so the the total number of arrangements is 24 x 2 = 48.
So, the probability that Childeric and Ethelred sit next to each other is 48/120 = 2/5.
Answer: D
_________________
Jeffery Miller Head of GMAT Instruction
GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions
Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
[#permalink]
Show Tags
02 Dec 2018, 12:52
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
60% (00:52) correct 
