It is currently 15 Nov 2018, 11:02
My Tests

Close

GMAT Club Daily Prep

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.

In a race, at least 3 and at most 5 runners will

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Moderator
Moderator
User avatar
Joined: 18 Apr 2015
Posts: 4876
Followers: 74

Kudos [?]: 973 [0], given: 4482

CAT Tests
In a race, at least 3 and at most 5 runners will [#permalink] New post 21 Aug 2017, 12:33
Expert's post
00:00

Question Stats:

64% (00:56) correct 35% (01:08) wrong based on 17 sessions


In a race, at least 3 and at most 5 runners will vie for gold, silver, and bronze medals. Which of the following could represent the total number of unique ways to distribute the three medals among the participants?

Indicate all such numbers.


❑ 3

❑ 6

❑ 12

❑ 24

❑ 30

❑ 60

[Reveal] Spoiler: OA
B,D,F
[Reveal] Spoiler: OA

_________________

Get the 2 FREE GREPrepclub Tests

1 KUDOS received
Director
Director
Joined: 03 Sep 2017
Posts: 521
Followers: 1

Kudos [?]: 327 [1] , given: 66

Re: In a race, at least 3 and at most 5 runners will [#permalink] New post 24 Sep 2017, 07:39
1
This post received
KUDOS
The crucial information to which paying attention is unique ways. Let's analyze the three occasions:
- if the runners are 3, there are 3! ways to allocate 3 medals among 3 people: 3*2*1 = 6 - Answer B
- if the runner are 4, we have to consider 4C3 to get the number of combinations without paying attention to the order, but since we are asked to count the unique ways, we have to multiply it by 3! in order to consider each variant as unique: 4C3*3!=4*3*2*1 = 24 - Answer D
- if runners are 5, the argument is above, we have to take 5C3*3! = 5!/(2!*3!)*3! = 10*3*2*1 = 60 - Answer F
1 KUDOS received
Intern
Intern
Joined: 17 Feb 2018
Posts: 31
Followers: 0

Kudos [?]: 24 [1] , given: 2

Re: In a race, at least 3 and at most 5 runners will [#permalink] New post 21 Feb 2018, 13:23
1
This post received
KUDOS
Each metal is different than the others.
Therefore, we can use 3P3,4P3 and 5P3 for the computation.
3p3=6 4p3=24 5p3=60
Re: In a race, at least 3 and at most 5 runners will   [#permalink] 21 Feb 2018, 13:23
Display posts from previous: Sort by

In a race, at least 3 and at most 5 runners will

  Question banks Downloads My Bookmarks Reviews Important topics  


GRE Prep Club Forum Home| About| Terms and Conditions and Privacy Policy| GRE Prep Club Rules| Contact

Powered by phpBB © phpBB Group

Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.