GMAT Changed on April 16th - Read about the latest changes here

It is currently 23 May 2018, 00:03

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

A company that ships boxes to a total of 12 distribution cen

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

Hide Tags

Expert Post
2 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 45256
A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 06 Mar 2014, 02:57
2
This post received
KUDOS
Expert's post
15
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

64% (01:10) correct 36% (01:19) wrong based on 738 sessions

HideShow timer Statistics

The Official Guide For GMAT® Quantitative Review, 2ND Edition

A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)

(A) 4
(B) 5
(C) 6
(D) 12
(E) 24

Problem Solving
Question: 132
Category: Arithmetic Elementary combinatorics
Page: 79
Difficulty: 600


GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!

_________________

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

Expert Post
3 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 45256
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 06 Mar 2014, 02:58
3
This post received
KUDOS
Expert's post
4
This post was
BOOKMARKED
SOLUTION

A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)

(A) 4
(B) 5
(C) 6
(D) 12
(E) 24

You can use trial and error method as well as algebraic approach:

Let # of colors needed be \(n\), then it must be true that \(n+C^2_n\geq{12}\) (\(C^2_n\) - # of ways to choose the pair of different colors from \(n\) colors when order doesn't matter) --> \(n+\frac{n(n-1)}{2}\geq{12}\) --> \(2n+n(n-1)\geq{24}\) --> \(n(n+1)\geq{24}\) --> as \(n\) is an integer (it represents # of colors) \(n\geq{5}\) --> \(n_{min}=5\).

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

4 KUDOS received
Manager
Manager
avatar
Joined: 04 Jan 2014
Posts: 120
GMAT 1: 660 Q48 V32
GMAT 2: 630 Q48 V28
GMAT 3: 680 Q48 V35
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 06 Mar 2014, 03:18
4
This post received
KUDOS
Let the colors be n. We need to find the min value of n for which nC2 >= 12.

Using answer choices, if n = 5; 5C2 = 10 so we are short by 2.
If n = 6, 6C2 = 15. Good!

Answer (C).
5 KUDOS received
Director
Director
User avatar
Joined: 25 Apr 2012
Posts: 702
Location: India
GPA: 3.21
WE: Business Development (Other)
Premium Member Reviews Badge
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 06 Mar 2014, 04:16
5
This post received
KUDOS
5
This post was
BOOKMARKED
A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)

(A) 4
(B) 5
(C) 6
(D) 12
(E) 24


Sol: Let's start with 4 minimum number of colors

So single color code we can make 4 nos.
Now if we need to make 2 color combination out of 4 we can do so in 4!/2!*2! or 4*3/2 or 6

So total we can make 4+6=10 color combinations but we have 12 boxes

So let's look at 5 we get 5 single color codes
and out 5 color choices, we can choose 2 in 5!/2!*3! ways or 10 ways.
So total we can have 5+10=15 color combinations.

So, minimum number we need will be 5

Ans is B........
_________________


“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

2 KUDOS received
Intern
Intern
avatar
Joined: 26 Feb 2012
Posts: 14
Concentration: Strategy, International Business
Schools: INSEAD Jan '16
GMAT 3: 640 Q49 V29
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 06 Mar 2014, 04:19
2
This post received
KUDOS
1
This post was
BOOKMARKED
Bunuel wrote:

A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)

(A) 4
(B) 5
(C) 6
(D) 12
(E) 24

Problem Solving
Question: 132
Category: Arithmetic Elementary combinatorics
Page: 79
Difficulty: 600



The statement says a single color or a pair of two different colors is chosen to represent each center.
Hence, the number of combinations have to be greater or equal than 12:

\(nC1 + nC2 >= 12\)

Where:

\(nC1 = n\)

\(nC2 = \frac{n*(n-1)}{2}\)

So:

\(nC1 + nC2 = n+\frac{n*(n-1)}{2} >= 12\)

\(\frac{2n+n*(n-1)}{2} >= 12\)

\(2n+n*(n-1) >= 24\)

\(n2+n >= 24\)

We can now pick values for n, which will be faster than solving:

If n=4,

\(n2+n = 20 < 24\)

If n=5,

\(n2+n = 29 >= 24\)

Answer: B (n=5)

For those willing to solve for n:

\(n2+n >= 24\)

\(n2+n -24 >0\)

Solving for n,

\(\frac{-1+-sqroot(1+96)}{2}\)

\(\frac{-1+-sqroot(97)}{2}\)


We don't really need to solve the square root:

- Negative option of the square root is not possible
- The positive option can be approximated by

\(\frac{1+sqroot(100)}{2} = (approx.)= 5\)

(but slightly less than 5)

Given that the inequality is:

\(n2+n -24 >0\)

Any value of n>=(slightly less than 5) will make the inequality positive.

Hence n=5

Answer: B (n=5)
Expert Post
2 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 45256
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 08 Mar 2014, 12:46
2
This post received
KUDOS
Expert's post
6
This post was
BOOKMARKED
Similar questions to practice:
each-student-at-a-certain-university-is-given-a-four-charact-151945.html
all-of-the-stocks-on-the-over-the-counter-market-are-126630.html
if-a-code-word-is-defined-to-be-a-sequence-of-different-126652.html
a-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html
a-5-digit-code-consists-of-one-number-digit-chosen-from-132263.html
a-company-that-ships-boxes-to-a-total-of-12-distribution-95946.html
a-company-plans-to-assign-identification-numbers-to-its-empl-69248.html
the-security-gate-at-a-storage-facility-requires-a-five-109932.html
all-of-the-bonds-on-a-certain-exchange-are-designated-by-a-150820.html
a-local-bank-that-has-15-branches-uses-a-two-digit-code-to-98109.html
a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html
baker-s-dozen-128782-20.html#p1057502
in-a-certain-appliance-store-each-model-of-television-is-136646.html
m04q29-color-coding-70074.html
john-has-12-clients-and-he-wants-to-use-color-coding-to-iden-107307.html
how-many-4-digit-even-numbers-do-not-use-any-digit-more-than-101874.html
a-certain-stock-exchange-designates-each-stock-with-a-85831.html
the-simplastic-language-has-only-2-unique-values-and-105845.html
m04q29-color-coding-70074.html
the-telephone-company-wants-to-add-an-area-code-composed-of-20252.html
_________________

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

1 KUDOS received
Manager
Manager
User avatar
Joined: 05 Jul 2015
Posts: 105
Concentration: Real Estate, International Business
GMAT 1: 600 Q33 V40
GPA: 3.3
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 22 Nov 2015, 11:48
1
This post received
KUDOS
I labeled the colors alphabetically and wrote them out

A
B
AB
C
CA,CB,
D
DA,DB,DC 4 Colors (ABCD) = 10 combinations so 1 more color will give more than 12 combinations.
Expert Post
EMPOWERgmat Instructor
User avatar
D
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11655
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 25 Nov 2015, 15:31
Hi DJ1986,

The 'brute force' approach that you used is PERFECT for these types of questions. When the answer choices are relatively small, it can sometimes be fastest/easiest to just put pen-to-pad and 'map out' all of the possibilities. In that way, you're not trying to make the solution overly-complicated and you're not starting at the screen (hoping that some idea will come to you). You'll likely find that you can take this approach on a few questions in the Quant section on Test Day, so don't be shy about using it (and practicing with it in mind).

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Director
Director
User avatar
G
Joined: 09 Mar 2016
Posts: 527
A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 25 Feb 2018, 08:10
Bunuel wrote:
SOLUTION

A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)

(A) 4
(B) 5
(C) 6
(D) 12
(E) 24

You can use trial and error method as well as algebraic approach:

Let # of colors needed be \(n\), then it must be true that \(n+C^2_n\geq{12}\) (\(C^2_n\) - # of ways to choose the pair of different colors from \(n\) colors when order doesn't matter) --> \(n+\frac{n(n-1)}{2}\geq{12}\) --> \(2n+n(n-1)\geq{24}\) --> \(n(n+1)\geq{24}\) --> as \(n\) is an integer (it represents # of colors) \(n\geq{5}\) --> \(n_{min}=5\).

Answer: B.


Bunuel - why are you adding \(n\) to \(C^2_n\geq{12}\) :? and not just write \(C^2_n\geq{12}\) without adding \(n\) ...
Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 45256
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 25 Feb 2018, 08:27
1
This post received
KUDOS
Expert's post
dave13 wrote:
Bunuel wrote:
SOLUTION

A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)

(A) 4
(B) 5
(C) 6
(D) 12
(E) 24

You can use trial and error method as well as algebraic approach:

Let # of colors needed be \(n\), then it must be true that \(n+C^2_n\geq{12}\) (\(C^2_n\) - # of ways to choose the pair of different colors from \(n\) colors when order doesn't matter) --> \(n+\frac{n(n-1)}{2}\geq{12}\) --> \(2n+n(n-1)\geq{24}\) --> \(n(n+1)\geq{24}\) --> as \(n\) is an integer (it represents # of colors) \(n\geq{5}\) --> \(n_{min}=5\).

Answer: B.


Bunuel - why are you adding \(n\) to \(C^2_n\geq{12}\) :? and not just write \(C^2_n\geq{12}\) without adding \(n\) ...


Highlighted the part that explains it. I'll let you figure out the rest.
_________________

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

Intern
Intern
avatar
B
Joined: 01 Oct 2016
Posts: 6
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 25 Feb 2018, 11:07
Nice one.
n + nC2 >= 12
n^2 + n -24>=0
Satisfies b = 5
Manager
Manager
avatar
S
Joined: 23 Sep 2016
Posts: 231
Premium Member Reviews Badge
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 26 Feb 2018, 04:14
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)

(A) 4
(B) 5
(C) 6
(D) 12
(E) 24

Problem Solving
Question: 132

Category: Arithmetic Elementary combinatorics
Page: 79
Difficulty: 600


GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!

IMO is B 12 distribution center then to save time we can plug in values lets take c as option then total colors are 6+6c2= only 6c2 will produce 15 colors then d and e is also eliminated lets try B as 5+5c2= 5+10=15 keep it as it is close to 12 and if we see option A then 4+4c2= 4+6=10 which is less than 12 colors so B is the answer
Expert Post
Target Test Prep Representative
User avatar
G
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 2611
Location: United States (CA)
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 27 Feb 2018, 10:20
Quote:

A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)

(A) 4
(B) 5
(C) 6
(D) 12
(E) 24m


Since we have only 12 distribution centers, we know we will need fewer than 12 different colors to identify them.

Let’s say we have 4 different colors; then 4C1 = 4 centers can be identified by one color and 4C2 = 6 centers can be identified by two different colors. So a total of 4 + 6 = 10 centers can be identified.

We see that if we have only 4 different colors, we don’t have enough ID codes to assign to the 12 centers. Therefore, we need one more color.

If we have 5 different colors, then 5C1 = 5 centers can be identified by one color, and 5C2 = 10 centers can be identified by two different colors. So a total of 5 + 10 = 15 centers can be identified.

We see that if we have 5 different colors, we have more than enough ID codes to assign to the 12 centers.

Answer: B
_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Director
Director
User avatar
G
Joined: 09 Mar 2016
Posts: 527
A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 23 Mar 2018, 10:52
Bunuel wrote:
SOLUTION

A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)

(A) 4
(B) 5
(C) 6
(D) 12
(E) 24

You can use trial and error method as well as algebraic approach:

Let # of colors needed be \(n\), then it must be true that \(n+C^2_n\geq{12}\) (\(C^2_n\) - # of ways to choose the pair of different colors from \(n\) colors when order doesn't matter) --> \(n+\frac{n(n-1)}{2}\geq{12}\) --> \(2n+n(n-1)\geq{24}\) --> \(n(n+1)\geq{24}\) --> as \(n\) is an integer (it represents # of colors) \(n\geq{5}\) --> \(n_{min}=5\).

Answer: B.


hello friends,

i have a few questions to understand the above solution in details :)

Why the inequily sign is less or equal 12 ? :? i understand there are 12 distribution centers, but we are looking the minimum number of combinations, if so than inequily sign should be less than 12 and not equal 12 no :?

Why are we dividing by 2 ? :?

can someone explain please ? :)

thanks!
BSchool Forum Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 2552
Location: India
GPA: 3.12
Premium Member CAT Tests
A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 23 Mar 2018, 12:21
1
This post was
BOOKMARKED
dave13 wrote:
Bunuel wrote:
SOLUTION

A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)

(A) 4
(B) 5
(C) 6
(D) 12
(E) 24

You can use trial and error method as well as algebraic approach:

Let # of colors needed be \(n\), then it must be true that \(n+C^2_n\geq{12}\) (\(C^2_n\) - # of ways to choose the pair of different colors from \(n\) colors when order doesn't matter) --> \(n+\frac{n(n-1)}{2}\geq{12}\) --> \(2n+n(n-1)\geq{24}\) --> \(n(n+1)\geq{24}\) --> as \(n\) is an integer (it represents # of colors) \(n\geq{5}\) --> \(n_{min}=5\).

Answer: B.


hello friends,

i have a few questions to understand the above solution in details :)

Why the inequily sign is less or equal 12 ? :? i understand there are 12 distribution centers, but we are looking the minimum number of combinations, if so than inequily sign should be less than 12 and not equal 12 no :?

Why are we dividing by 2 ? :?

can someone explain please ? :)

thanks!



Hi dave13

The question is asking for the minimum number of colors needed to identify 12 distribution centers.
We are told that a single color or a pair of two different colors is chosen to represent each center.

Here, we are assuming the number of colors to be n

Since the combination of 2 colors is also a viable means to color code the distribution centres, we
can write \(n+C^2_n\geq{12}\)

Here, \(C^2_n = \frac{n!}{(n-2)!*2!} = \frac{n(n-1)(n-2)!}{(n-2)!*2} = \frac{n(n-1)}{2}\) (Cancelling out common terms)

We are substituting this value of \(C^2_n\) in \(n+C^2_n\geq{12}\) which brings us to

\(n+\frac{n(n-1)}{2}\geq{12}\) -> \(2n+n(n-1)\geq{24}\) which can be further simplified as \(n(n+1)\geq{24}\)

The number is greater than 24, if and only if n=5(at a number lower than it - n=4, we get \(n(n+1) = 20\)

Hope it is clearer now!
_________________

You've got what it takes, but it will take everything you've got

Director
Director
User avatar
G
Joined: 09 Mar 2016
Posts: 527
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 23 Mar 2018, 13:21
pushpitkc wrote:
dave13 wrote:
Bunuel wrote:
SOLUTION

A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)

(A) 4
(B) 5
(C) 6
(D) 12
(E) 24

You can use trial and error method as well as algebraic approach:

Let # of colors needed be \(n\), then it must be true that \(n+C^2_n\geq{12}\) (\(C^2_n\) - # of ways to choose the pair of different colors from \(n\) colors when order doesn't matter) --> \(n+\frac{n(n-1)}{2}\geq{12}\) --> \(2n+n(n-1)\geq{24}\) --> \(n(n+1)\geq{24}\) --> as \(n\) is an integer (it represents # of colors) \(n\geq{5}\) --> \(n_{min}=5\).

Answer: B.


hello friends,

i have a few questions to understand the above solution in details :)

Why the inequily sign is less or equal 12 ? :? i understand there are 12 distribution centers, but we are looking the minimum number of combinations, if so than inequily sign should be less than 12 and not equal 12 no :?

Why are we dividing by 2 ? :?

can someone explain please ? :)

thanks!



Hi dave13

The question is asking for the minimum number of colors needed to identify 12 distribution centers.
We are told that a single color or a pair of two different colors is chosen to represent each center.

Here, we are assuming the number of colors to be n

Since the combination of 2 colors is also a viable means to color code the distribution centres, we
can write \(n+C^2_n\geq{12}\)

Here, \(C^2_n = \frac{n!}{(n-2)!*2!} = \frac{n(n-1)(n-2)!}{(n-2)!*2} = \frac{n(n-1)}{2}\) (Cancelling out common terms)

We are substituting this value of \(C^2_n\) in \(n+C^2_n\geq{12}\) which brings us to

\(n+\frac{n(n-1)}{2}\geq{12}\) -> \(2n+n(n-1)\geq{24}\) which can be further simplified as \(n(n+1)\geq{24}\)

The number is greater than 24, if and only if n=5(at a number lower than it - n=4, we get \(n(n+1) = 20\)

Hope it is clearer now!


Hi pushpitkc

thank you so much for taking time to explain . highly apprecited :)

everything is clear but this one

\(\frac{n(n-1)(n-2)!}{(n-2)!*2}\)

how in numerator do you get n(n-1)(n-2) :?

what does it mean ? logic behind? whats explanation :) ?

thank you :-)
BSchool Forum Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 2552
Location: India
GPA: 3.12
Premium Member CAT Tests
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 23 Mar 2018, 13:40
Hi dave13

5!(whose value is 120) can be rewritten as 5*4*3!(whose value is 20*6 = 120)

Similarly n! is otherwise written as (n)(n-1)(n-2)!

Hope that helps!
_________________

You've got what it takes, but it will take everything you've got

Director
Director
User avatar
G
Joined: 09 Mar 2016
Posts: 527
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 23 Mar 2018, 13:57
pushpitkc wrote:
Hi dave13

5!(whose value is 120) can be rewritten as 5*4*3!(whose value is 20*6 = 120)

Similarly n! is otherwise written as (n)(n-1)(n-2)!

Hope that helps!



pushpitkc But 5! is not mentioned at that stage of solving ... :? you get 5 only ...only in the end :? is it example that you put forward :? or what got confused...
1 KUDOS received
BSchool Forum Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 2552
Location: India
GPA: 3.12
Premium Member CAT Tests
A company that ships boxes to a total of 12 distribution cen [#permalink]

Show Tags

New post 23 Mar 2018, 14:09
1
This post received
KUDOS
dave13
I was trying to illustrate by means of an example by considering n=5

The formula for a combination is
\(C^2_n = \frac{n!}{(n-2)!*2!}\)

Since we are given n! and it becomes n(n-1)(n-2)! and the (n-2)! in the numerator and denominator get cancelled.

After that, what you are left with is (n)(n-1)/2

Hope it is clear now.
_________________

You've got what it takes, but it will take everything you've got

A company that ships boxes to a total of 12 distribution cen   [#permalink] 23 Mar 2018, 14:09
Display posts from previous: Sort by

A company that ships boxes to a total of 12 distribution cen

  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®.