It is currently 13 Dec 2017, 20:40

Decision(s) Day!:

CHAT Rooms | Ross R1 | Kellogg R1 | Darden R1 | Tepper R1


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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

In a room are five chairs to accommodate 3 people, one

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

1 KUDOS received
Intern
Intern
avatar
Joined: 03 Dec 2013
Posts: 16

Kudos [?]: 13 [1], given: 57

Location: Uzbekistan
Concentration: Finance, Entrepreneurship
GMAT 1: 620 Q42 V33
GRE 1: 600 Q790 V400
GPA: 3.4
WE: Analyst (Commercial Banking)
GMAT ToolKit User Premium Member Reviews Badge
In a room are five chairs to accommodate 3 people, one [#permalink]

Show Tags

New post 12 Mar 2014, 01:46
1
This post received
KUDOS
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

79% (00:46) correct 21% (01:00) wrong based on 132 sessions

HideShow timer Statistics

In a room are five chairs to accommodate 3 people, one person to a chair. How many seating arrangements are possible?

A. 45
B. 60
C. 72
D. 90
E. 120

I found this problem tricky.

If you like this problem, please, be generous to give a Kudo.
[Reveal] Spoiler: OA

Kudos [?]: 13 [1], given: 57

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42597

Kudos [?]: 135562 [0], given: 12699

Re: In a room are five chairs to accommodate 3 people, one [#permalink]

Show Tags

New post 12 Mar 2014, 03:12
aja1991 wrote:
In a room are five chairs to accommodate 3 people, one person to a chair. How many seating arrangements are possible?

A. 45
B. 60
C. 72
D. 90
E. 120

I found this problem tricky.

If you like this problem, please, be generous to give a Kudo.


\(C^3_5*3!=60\), where \(C^3_5\) is the number of ways to choose 3 chairs for 3 people, and 3! is the number of arrangements of 3 people on those chairs.

Or directly: \(P^3_5=60\): choosing 3 out of 5, where order matters.

Answer: B.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 135562 [0], given: 12699

Intern
Intern
avatar
Joined: 21 Jan 2014
Posts: 14

Kudos [?]: 8 [0], given: 9

Location: Germany
Concentration: Entrepreneurship, Strategy
GMAT Date: 05-28-2014
GPA: 3.9
WE: Consulting (Entertainment and Sports)
Re: In a room are five chairs to accommodate 3 people, one [#permalink]

Show Tags

New post 12 Mar 2014, 14:16
I actually solved this problem by the following thoughts (derived from factorials):
First Person: 5 Chairs to choose
Second Person: 4 Chairs to choose
Third Person: 3 Chairs to choose
Then I simply multiplied this values and got the answer. Basically it’s 5! – 2!.

Now the interesting part: I just did that because I had no other idea, how to tackle this problem. Would someone please tell me, if I was just lucky? (and probably explain it differently)
Unfortunately I can’t understand what you are exactly calculating in your solution Bunuel.

Kudos [?]: 8 [0], given: 9

Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
G
Joined: 16 Oct 2010
Posts: 7793

Kudos [?]: 18119 [1], given: 236

Location: Pune, India
Re: In a room are five chairs to accommodate 3 people, one [#permalink]

Show Tags

New post 12 Mar 2014, 23:29
1
This post received
KUDOS
Expert's post
holdem wrote:
I actually solved this problem by the following thoughts (derived from factorials):
First Person: 5 Chairs to choose
Second Person: 4 Chairs to choose
Third Person: 3 Chairs to choose
Then I simply multiplied this values and got the answer. Basically it’s 5! – 2!.

Now the interesting part: I just did that because I had no other idea, how to tackle this problem. Would someone please tell me, if I was just lucky? (and probably explain it differently)
Unfortunately I can’t understand what you are exactly calculating in your solution Bunuel.


You used the Basic Counting Principle which is absolutely fine.
Answer is 5*4*3 = 60 (which by the way, is not the same as 5! - 2! = 118)

Here are some discussions on Basic Counting Principle and more:
http://www.veritasprep.com/blog/2011/10 ... inatorics/
http://www.veritasprep.com/blog/2011/10 ... ts-part-i/
http://www.veritasprep.com/blog/2011/10 ... s-part-ii/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Kudos [?]: 18119 [1], given: 236

Intern
Intern
avatar
B
Joined: 07 Aug 2017
Posts: 3

Kudos [?]: 0 [0], given: 4

Re: In a room are five chairs to accommodate 3 people, one [#permalink]

Show Tags

New post 25 Nov 2017, 01:56
Bunuel wrote:
aja1991 wrote:
In a room are five chairs to accommodate 3 people, one person to a chair. How many seating arrangements are possible?

A. 45
B. 60
C. 72
D. 90
E. 120

I found this problem tricky.

If you like this problem, please, be generous to give a Kudo.


\(C^3_5*3!=60\), where \(C^3_5\) is the number of ways to choose 3 chairs for 3 people, and 3! is the number of arrangements of 3 people on those chairs.

Or directly: \(P^3_5=60\): choosing 3 out of 5, where order matters.

Answer: B.



-------------------------------------------------------------------------------------------------------------------------------------------------------------
Why the two empty chairs are not accounted for? I mean why it's not \(C^3_5*3!*2!\) , where 2! is the empty chairs. Because the three person can have 2 adjacent empty chairs in between them or no empty chair or alternate empty chairs.

Kudos [?]: 0 [0], given: 4

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42597

Kudos [?]: 135562 [0], given: 12699

Re: In a room are five chairs to accommodate 3 people, one [#permalink]

Show Tags

New post 25 Nov 2017, 03:14
GMATMBA5 wrote:
Bunuel wrote:
aja1991 wrote:
In a room are five chairs to accommodate 3 people, one person to a chair. How many seating arrangements are possible?

A. 45
B. 60
C. 72
D. 90
E. 120

I found this problem tricky.

If you like this problem, please, be generous to give a Kudo.


\(C^3_5*3!=60\), where \(C^3_5\) is the number of ways to choose 3 chairs for 3 people, and 3! is the number of arrangements of 3 people on those chairs.

Or directly: \(P^3_5=60\): choosing 3 out of 5, where order matters.

Answer: B.



-------------------------------------------------------------------------------------------------------------------------------------------------------------
Why the two empty chairs are not accounted for? I mean why it's not \(C^3_5*3!*2!\) , where 2! is the empty chairs. Because the three person can have 2 adjacent empty chairs in between them or no empty chair or alternate empty chairs.


I don;t understand what you mean. The chairs are standing in a row. We are accommodating 3 people.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 135562 [0], given: 12699

Intern
Intern
avatar
B
Joined: 07 Aug 2017
Posts: 3

Kudos [?]: 0 [0], given: 4

Re: In a room are five chairs to accommodate 3 people, one [#permalink]

Show Tags

New post 25 Nov 2017, 04:47
-------------------------------------------------------------------------------------------------------------------------------------------------------------
Why the two empty chairs are not accounted for? I mean why it's not \(C^3_5*3!*2!\) , where 2! is the empty chairs. Because the three person can have 2 adjacent empty chairs in between them or no empty chair or alternate empty chairs.[/quote]

I don;t understand what you mean. The chairs are standing in a row. We are accommodating 3 people.[/quote]


------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Suppose P1C, P2C and P3C denote three persons sitting on three different chairs and EC denotes an empty chair.

With \(C^3_5*3!\), we determined the pattern of three persons sitting on 5 chairs, with interchanging seats.
But is the seating arrangement P1C-EC-P2C-EC-P3C not different from P1C-P2C-EC-EC-P3C. Do we not have to consider empty chairs as well for forming the pattern or is it that which chairs are left empty doesn't matter.

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Kudos [?]: 0 [0], given: 4

Expert Post
1 KUDOS received
Math Expert
User avatar
D
Joined: 02 Aug 2009
Posts: 5347

Kudos [?]: 6120 [1], given: 121

In a room are five chairs to accommodate 3 people, one [#permalink]

Show Tags

New post 25 Nov 2017, 05:00
1
This post received
KUDOS
Expert's post
Quote:
Quote:
GMATMBA5 wrote:
-------------------------------------------------------------------------------------------------------------------------------------------------------------
Why the two empty chairs are not accounted for? I mean why it's not \(C^3_5*3!*2!\) , where 2! is the empty chairs. Because the three person can have 2 adjacent empty chairs in between them or no empty chair or alternate empty chairs.


I don;t understand what you mean. The chairs are standing in a row. We are accommodating 3 people.



------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Suppose P1C, P2C and P3C denote three persons sitting on three different chairs and EC denotes an empty chair.

With \(C^3_5*3!\), we determined the pattern of three persons sitting on 5 chairs, with interchanging seats.
But is the seating arrangement P1C-EC-P2C-EC-P3C not different from P1C-P2C-EC-EC-P3C. Do we not have to consider empty chairs as well for forming the pattern or is it that which chairs are left empty doesn't matter.

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------



Hi...

when you are choosing 3 chairs out of 5 by 5C3, you are already catering for where the E are ..
so FFFEE is taken as a different scenario than FFEEF....
Now these three can be filled in 3! ways that is why 5C3*3!

let three person be A, B and C
here 3!=3*2=6 means FFFEE is
ABCEE
ACBEE
BACEE
BCAEE
CABEE
CBAEE

hope it helps
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6120 [1], given: 121

In a room are five chairs to accommodate 3 people, one   [#permalink] 25 Nov 2017, 05:00
Display posts from previous: Sort by

In a room are five chairs to accommodate 3 people, one

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.