Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 24 Jun 2016, 20:30

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

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

4 chairs are there

Author Message
Intern
Joined: 25 Mar 2008
Posts: 43
Followers: 0

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

Show Tags

26 Nov 2009, 09:11
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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".

CEO
Joined: 17 Nov 2007
Posts: 3580
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 484

Kudos [?]: 2794 [0], given: 359

Re: 4 chairs are there [#permalink]

Show Tags

26 Nov 2009, 09:47
Expert's post
There a few ways how to approach this problem:

1) Enumeration of possibilities

ABxx
AxBx
AxxB
xABx
xAxB
xxAB

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

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
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Intern
Joined: 25 Mar 2008
Posts: 43
Followers: 0

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

Re: 4 chairs are there [#permalink]

Show Tags

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

CEO
Joined: 17 Nov 2007
Posts: 3580
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 484

Kudos [?]: 2794 [0], given: 359

Re: 4 chairs are there [#permalink]

Show Tags

26 Nov 2009, 10:23
Expert's post
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.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Joined: 20 Aug 2009
Posts: 311
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
Followers: 16

Kudos [?]: 128 [0], given: 69

Re: 4 chairs are there [#permalink]

Show Tags

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"
Intern
Joined: 25 Mar 2008
Posts: 43
Followers: 0

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

Re: 4 chairs are there [#permalink]

Show Tags

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.

CEO
Joined: 17 Nov 2007
Posts: 3580
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 484

Kudos [?]: 2794 [0], given: 359

Re: 4 chairs are there [#permalink]

Show Tags

26 Nov 2009, 22:54
Expert's post
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.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Intern
Joined: 25 Mar 2008
Posts: 43
Followers: 0

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

Re: 4 chairs are there [#permalink]

Show Tags

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 ?
Manager
Joined: 19 Nov 2007
Posts: 225
Followers: 1

Kudos [?]: 207 [0], given: 1

Re: 4 chairs are there [#permalink]

Show Tags

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.
Manager
Joined: 19 Nov 2007
Posts: 225
Followers: 1

Kudos [?]: 207 [0], given: 1

Re: 4 chairs are there [#permalink]

Show Tags

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
Re: 4 chairs are there   [#permalink] 28 Nov 2009, 07:07
Display posts from previous: Sort by