NEW HERE? LEARN MORE
closeclose
Last visit was: 23 Apr 2024, 15:55 It is currently 23 Apr 2024, 15:55
Decision Tracker
My Rewards
New posts
New comers' posts

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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

4 chairs are there

User avatar
ghentu
Intern
Intern
Joined: 25 Mar 2008
Posts: 29
Own Kudos [?]: 17 [0]
Given Kudos: 0
Send PM
4 chairs are there [#permalink] New post   26 Nov 2009, 09:11
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Hide Show timer Statistics
4 chairs are there. Two person A and B wants to sit there such that A is always left to B. How many ways it can be done ?



Can we write it as 4C2 way ? I am doubtful as there is a constraint " A is always left to B".

whats your comments ?



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
User avatar
B
walker
SVP
SVP
Joined: 17 Nov 2007
Posts: 2408
Own Kudos [?]: 10035 [0]
Given Kudos: 361
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Send PM
GMAT ToolKit User
Re: 4 chairs are there [#permalink] New post   26 Nov 2009, 09:47
Expert Reply
There a few ways how to approach this problem:

1) Enumeration of possibilities

ABxx
AxBx
AxxB
xABx
xAxB
xxAB

Answer: 6

2)
A at 1st left place: C^3_1 = 3 ways for B
A at 2st left place: C^2_1 = 2 ways for B
A at 3st left place: C^1_1 = 1 ways for B

Answer: 6

3) trick: symmetry

P^4_2 = 12 - the total number of ways
A is always left to B in exactly 1/2 of all cases.

Answer: 1/2 * P^4_2 = 6
User avatar
ghentu
Intern
Intern
Joined: 25 Mar 2008
Posts: 29
Own Kudos [?]: 17 [0]
Given Kudos: 0
Send PM
Re: 4 chairs are there [#permalink] New post   26 Nov 2009, 10:11
instead of doing your way , can i just use the one liner combination rule ?

i.e pick 2 chairs out of 4 ==> 4C2 = 6

As you see I also got the same answer as you got ....i just applied the combination definition rule

But is it a valid math ? That was intent of this question .
If its not valid math , then explain why its not
User avatar
B
walker
SVP
SVP
Joined: 17 Nov 2007
Posts: 2408
Own Kudos [?]: 10035 [0]
Given Kudos: 361
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Send PM
GMAT ToolKit User
Re: 4 chairs are there [#permalink] New post   26 Nov 2009, 10:23
Expert Reply
if you apply 4C2, you need to understand why. 4C2 means how many ways you have to choose 2 chairs out of 4 chairs but it is not enough.

4C2 is right if you think this way:

1) let's choose any 2 seats out of 4.
2) For chosen 2 seats we pick left seat for A and right seat for B.
User avatar
shalva
Tuck School Moderator
Joined: 20 Aug 2009
Posts: 203
Own Kudos [?]: 330 [0]
Given Kudos: 69
Location: Tbilisi, Georgia
Schools:Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
 Q50  V47
Send PM
Re: 4 chairs are there [#permalink] New post   26 Nov 2009, 13:14
walker wrote:
3) trick: symmetry

P^4_2 = 12 - the total number of ways
A is always left to B in exactly 1/2 of all cases.

Answer: 1/2 * P^4_2 = 6


I think this is the best way.

Variation without P and combinatorial formulas:

"A" can pick any of 4 chairs. In each case there would be 3 other chairs "B" could sit on.

4*3=12 - # of total possibilities. 12/2=6 - # of ways they sit when "A" is left to "B"
User avatar
ghentu
Intern
Intern
Joined: 25 Mar 2008
Posts: 29
Own Kudos [?]: 17 [0]
Given Kudos: 0
Send PM
Re: 4 chairs are there [#permalink] New post   26 Nov 2009, 19:25
walker wrote:
if you apply 4C2, you need to understand why. 4C2 means how many ways you have to choose 2 chairs out of 4 chairs but it is not enough.

4C2 is right if you think this way:

1) let's choose any 2 seats out of 4.
2) For chosen 2 seats we pick left seat for A and right seat for B.


but that red colored constraint is a violation on the very definition of combination ....definition of combination talks about group ...it does not bother about specific order....is not it ? ....so, we really cant write down 4C2 here as per the combination rule .

I am happy with your earlier solution.

I found in a book ...they used 4C2 in the solution . I object this ...because this is against the definition of combination ( as there is a constraint).

But I observe you did not use this 4C2 thingies and hence I liked your solution.

comments please.
User avatar
B
walker
SVP
SVP
Joined: 17 Nov 2007
Posts: 2408
Own Kudos [?]: 10035 [0]
Given Kudos: 361
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Send PM
GMAT ToolKit User
Re: 4 chairs are there [#permalink] New post   26 Nov 2009, 22:54
Expert Reply
4C2 means the number of ways to choose 2 objects out of 4. No more, no less. So, nothing is wrong with using 4C2 in my approach.
User avatar
ghentu
Intern
Intern
Joined: 25 Mar 2008
Posts: 29
Own Kudos [?]: 17 [0]
Given Kudos: 0
Send PM
Re: 4 chairs are there [#permalink] New post   28 Nov 2009, 00:51
walker wrote:
4C2 means the number of ways to choose 2 objects out of 4. No more, no less. So, nothing is wrong with using 4C2 in my approach.


what if the question asked was

4 chairs are there. Two person A and B wants to sit there . How many ways it can be done ?

//note i removed the clause "such that A is always left to B".

is it still the same answer 4C2 ? does that clause have no impact at all ? why ?
User avatar
jade3
Manager
Manager
Joined: 19 Nov 2007
Posts: 97
Own Kudos [?]: 799 [0]
Given Kudos: 1
Send PM
Re: 4 chairs are there [#permalink] New post   28 Nov 2009, 07:06
There are four places and two of them want to sit in only one way. If we combine two (A & B) places into one, then there are 6 ways to occupy three places and 1 way to arrange the A and B.

The total ways is 6*1=6.
User avatar
jade3
Manager
Manager
Joined: 19 Nov 2007
Posts: 97
Own Kudos [?]: 799 [0]
Given Kudos: 1
Send PM
Re: 4 chairs are there [#permalink] New post   28 Nov 2009, 07:07
ghentu wrote:
walker wrote:
4C2 means the number of ways to choose 2 objects out of 4. No more, no less. So, nothing is wrong with using 4C2 in my approach.


what if the question asked was

4 chairs are there. Two person A and B wants to sit there . How many ways it can be done ?

//note i removed the clause "such that A is always left to B".

is it still the same answer 4C2 ? does that clause have no impact at all ? why ?


Then the answer would be 6*2=12 ways

This is an arrangement and not a selection. IMO there is no need to use Combination formula



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
GMAT Club Bot
gmatclubot
Re: 4 chairs are there [#permalink]
28 Nov 2009, 07:07
Moderators:
Bunuel
Math Expert
92883 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions