Sep 19 08:00 PM EDT  09:00 PM EDT Learn how to separate yourself from the competition during this presentation led by admissions expert and Accepted CEO Linda Abraham as she lays out a proven framework for standout applications, common mistakes to avoid, and how to mitigate low stats. Sep 19 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GRE. Limited for the first 99 registrants. Sep 24 08:00 AM PDT  09:00 AM PDT Join a free 1hour webinar and learn how to create the ultimate study plan that will lead you to a high GRE score, and be accepted to the upcoming winter deadlines. Save your spot today! Tuesday, September 24th at 8 AM PT Sep 25 08:00 PM PDT  09:00 PM PDT Regardless of whether you choose to study with Greenlight Test Prep, I believe you'll benefit from my many free resources. Sep 27 08:00 PM PDT  09:00 PM PDT Sign up for 1 Week GRE Prep Sep 28 08:00 PM PDT  11:00 PM PDT We believe that our approach to online instruction provides a more intuitive experience than anything else in the online GRE prep market. Sep 28 08:00 PM PDT  09:00 PM PDT Friends, come hang with some Admissionado consultants, in person, at the QS Tour. Lisa Pearo (HBS) will be in Boston on the 24th, and Seth Shapiro (HBS) and Doris Huang (Wharton) will be in NYC on the 28th. Come chat us up. We’ll straighten you out!
Author 
Message 
TAGS:


Founder
Joined: 18 Apr 2015
Posts: 8120
Followers: 157
Kudos [?]:
1700
[0], given: 7473

Four pool balls—A, B, C, D—are randomly arranged in a straig [#permalink]
08 May 2019, 04:28
Question Stats:
57% (00:26) correct
42% (00:40) wrong based on 7 sessions
Four pool balls—A, B, C, D—are randomly arranged in a straight line. What is the probability that the order will actually be A, B, C, D ? (A) \(\frac{1}{4}\) (B) \(\frac{1}{4C_4}\) (C) \(\frac{1}{4P_4}\) (D) \(\frac{1}{2!}\) (E) \(\frac{1}{3!}\)
_________________
Get the 2 FREE GREPrepclub Tests




Moderator
Joined: 03 Apr 2019
Posts: 143
Followers: 2
Kudos [?]:
111
[0], given: 2

Re: Four pool balls—A, B, C, D—are randomly arranged in a straig [#permalink]
12 May 2019, 20:14
Can someone please help me with this question?



Founder
Joined: 18 Apr 2015
Posts: 8120
Followers: 157
Kudos [?]:
1700
[0], given: 7473

Re: Four pool balls—A, B, C, D—are randomly arranged in a straig [#permalink]
13 May 2019, 02:18
A ball cannot exist in two slots, so repetition is not allowed. Each ball is given a different identity A, B, C, and D, so there are no indistinguishable objects. Here, n = 4 (number of balls to arrange) in r = 4 (positions). We know the problem type, and the formula to use. Hence, by Formula 2, the number of arrangements possible is \(4P_4\) , and {A, B, C, D} is just one of the arrangements. Hence, the probability is 1 in \(4P_4\) , or \(\frac{1}{4P_4}\). The answer is (C).
_________________
Get the 2 FREE GREPrepclub Tests



MyGuru Representative
Affiliations: Partner at MyGuru LLC.
Joined: 13 May 2019
Posts: 130
Location: United States
GMAT 1: 770 Q51 V44
WE: Education (Education)
Followers: 3
Kudos [?]:
125
[1]
, given: 2

Re: Four pool balls—A, B, C, D—are randomly arranged in a straig [#permalink]
13 May 2019, 07:09
1
This post received KUDOS
An easy mnemonic way to distinguish between Permutations and Combinations is that a Permutation is a permanent unique iteration where order matters, such as this situation where A  B  C  D is not the same outcome as C B  D  A. Whereas, a Combination is an iteration where order does not matter and can be changed for instance if we were to select four candidates for two identical job openings this would be a combination since it would not matter whether a person were selected for opening one or two. As it pertains to this specific question, we should be able to conceptually determine that there is only one method for selecting exactly A  B  C  D as requested. Then, we can determine that there would be four ways to select the first ball, three for the second ball, two for the third ball, and only one for the fourth ball. Since we are selecting each of those four balls, it is necessary to multiply each of the possible outcomes together to determine the total number of possibilities. Therefore, there would be 4 x 3 x 2 x 1 or 4! = 16 total possibilities and we are exactly seeking one of those choices, so you may be able to determine that the correct answer should be 1/16. However, of course neither 1/16 nor 1/4! are available options, so we just recall our mnemonic to determine that this scenario is a Permutation that would be permanently set without the flexibility to change spots to inform that we would need to select the P formula in choice C rather than the C formula in choice B.
_________________
Stefan Maisnier Director of Online Tutoring, MyGuru www.myguruedge.com Sign up today for a free 30minute online GRE tutoring session! Like us on Facebook Watch us on YouTube Follow us on Twitter Follow us on Linkedin




Re: Four pool balls—A, B, C, D—are randomly arranged in a straig
[#permalink]
13 May 2019, 07:09





