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

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A company that ships boxes to a total of 12 distribution [#permalink]

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16 Jun 2010, 09:23
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62% (01:03) correct 38% (01:14) wrong based on 791 sessions

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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 colors in a pair does not matter)

A. 4
B. 5
C. 6
D. 12
E. 24
[Reveal] Spoiler: OA

Last edited by Bunuel on 31 Jul 2012, 13:49, edited 1 time in total.
Edited the question and added the OA.

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Re: need help to solve math question [#permalink]

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16 Jun 2010, 10:12
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This is a combination problem that can be formulated in the following matter:
XC1 + XC2 >= 12

Lets try answer 4. We will get 4C1+4C2= 4+6 which is less than 10.
Trying 5, we will get 5C1+5C2= 5+10 which is greater than 12,

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Re: need help to solve math question [#permalink]

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16 Jun 2010, 10:20
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chintzzz 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 colors in a pair does not matter)
A)4
B)5
C)6
D)12
E)24

You can solve by trial and error or use algebra.

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) $$n\geq{5}$$ --> $$n_{min}=5$$.

Hope it's clear.
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Re: A company that ships boxes to a total of 12 distribution [#permalink]

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19 Dec 2013, 06:40
I tried to do it with writing the possibilities out:

B(Blue) R(Red) Y(Yellow) P(Pink)

B
BR
RB
R
Y
YR
YB
RY
BY
P
PR
RP
PB
BP
PY
YP

I already reach 16 different combinations with only 4 colours, but the OA is 5? what´s my mistake?

EDIT:

Just figured that the ordering does not count as 2 different orders.. therefore, we need 5 colours.. thanks anyway

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Re: A company that ships boxes to a total of 12 distribution [#permalink]

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19 Dec 2013, 08:10
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Re: A company that ships boxes to a total of 12 distribution [#permalink]

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18 Feb 2015, 15:23
Can anyone tell me when we use n^2 and when we use n+nC2 ????

In this color question we use n+nC2 >= 12

In integer questions we use n^2>=15 ....

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Re: A company that ships boxes to a total of 12 distribution [#permalink]

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18 Feb 2015, 22:24
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Hi gmathopeful90,

The "restrictions" in the question are what dictate the math.

Consider these possible scenarios:

1) You have 5 different colors to choose from and two different rooms to paint. You can use the same color in both rooms. How many different color combinations are there for the two rooms?

Here, the first room could be 5 different colors and the second room could be 5 different colors, so (5)(5) = 5^2 = 25 options.

2) You have 5 different colors to choose from and two different rooms to paint. You CANNOT use the same color in both rooms. How many different color combinations are there for the two rooms?

Here, the first room could be 5 different colors; once you assign that first color, the second room could only be 4 different colors, so (5)(4) = 20 options.

3) You have 5 different colors to choose from. How many different 1-color and 2-color codes can you form with the following restrictions: the 2-color codes must use 2 DIFFERENT colors and the order of the colors does not matter (so blue-green is the SAME code as green-blue)?

Here, you start with the 5 different 1-color codes, then 5c2 different 2-color codes = 5 + 10 = 15 codes.

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# Rich Cohen

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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!*********************** Kudos [?]: 3505 [1], given: 173 Manager Joined: 06 Dec 2014 Posts: 67 Kudos [?]: 1 [0], given: 4 GMAT 1: 670 Q48 V34 Re: A company that ships boxes to a total of 12 distribution [#permalink] ### Show Tags 19 Feb 2015, 13:59 Thanks for the reply Do you mean in questions where we assume order of colors in cominations matters, we can use 5*4.. But where color doesn't matter, we use 5C2 ??? This explains stuff for me Kudos [?]: 1 [0], given: 4 EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 10111 Kudos [?]: 3505 [1], given: 173 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 [#permalink] ### Show Tags 19 Feb 2015, 14:24 1 This post received KUDOS Expert's post 2 This post was BOOKMARKED Hi gmathopeful90, You've hit on THE key difference between Permutation and Combination questions: does the order MATTER or not. IF you're putting things in order (the word "arrange" or "arrangements" often shows up in these types of questions), then you have to keep track of the number of options at each "step" and standard multiplication is involved. IF you're picking combinations of things (the word "combination" is the common word in these questions), then the order of the items does NOT matter and you have to use the Combination Formula. One of the interesting "design elements" of Official GMAT questions is that you can use either of the above approaches on certain types of prompts - you just have to be careful about how you set up the math (and you have to be really organized with your work). 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
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Re: A company that ships boxes to a total of 12 distribution [#permalink]

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12 Jul 2016, 21:14
I am confused with the line 'order doesn't matter'.

Does it mean BR = RB or BR =/ to RB?.

ij78cp wrote:
I tried to do it with writing the possibilities out:

B(Blue) R(Red) Y(Yellow) P(Pink)

B
BR
RB
R
Y
YR
YB
RY
BY
P
PR
RP
PB
BP
PY
YP

I already reach 16 different combinations with only 4 colours, but the OA is 5? what´s my mistake?

EDIT:

Just figured that the ordering does not count as 2 different orders.. therefore, we need 5 colours.. thanks anyway

Kudos [?]: 7 [0], given: 79

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
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GRE 1: 340 Q170 V170
Re: A company that ships boxes to a total of 12 distribution [#permalink]

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14 Jul 2016, 07:06
Hi ameyaprabhu,

When the order doesn't matter, RB and BR are the SAME option (so you can't count it twice, you can only count it once). In these sorts of questions, it can often be fastest to just 'list out' the possibilities (as opposed to doing lots of complex calculations).

GMAT assassins aren't born, they're made,
Rich
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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
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www.empowergmat.com/

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Re: A company that ships boxes to a total of 12 distribution [#permalink]

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22 Feb 2017, 02:41
Hi writing the possibilities should be like this:

1-color code: A, B, C, D, E --> 5 POS.
2-color code: AB, AC, BC, DA, DB, DC, EA, EB, EC, ED --> 10 POS.

ij78cp wrote:
I tried to do it with writing the possibilities out:

B(Blue) R(Red) Y(Yellow) P(Pink)

B
BR
RB
R
Y
YR
YB
RY
BY
P
PR
RP
PB
BP
PY
YP

I already reach 16 different combinations with only 4 colours, but the OA is 5? what´s my mistake?

EDIT:

Just figured that the ordering does not count as 2 different orders.. therefore, we need 5 colours.. thanks anyway

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Re: A company that ships boxes to a total of 12 distribution [#permalink]

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24 Apr 2017, 23:50
Bunuel wrote:
chintzzz 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 colors in a pair does not matter)
A)4
B)5
C)6
D)12
E)24

You can solve by trial and error or use algebra.

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) $$n\geq{5}$$ --> $$n_{min}=5$$.

Hope it's clear.

Could you please explain me how you get [fraction]n(n-1)/2 from C^2_n? Shouldn't it be [fraction]n!/k!(n-k)! ?
Thanks

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Re: A company that ships boxes to a total of 12 distribution [#permalink]

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25 Apr 2017, 02:04
matteogr wrote:
Bunuel wrote:
chintzzz 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 colors in a pair does not matter)
A)4
B)5
C)6
D)12
E)24

You can solve by trial and error or use algebra.

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) $$n\geq{5}$$ --> $$n_{min}=5$$.

Hope it's clear.

Could you please explain me how you get n(n-1)/2 from C^2_n? Shouldn't it be n!/k!(n-k)! ?
Thanks

$$C^2_n=\frac{n!}{(n-2)!*2!}=\frac{(n-2)!*(n-1)*n}{(n-2)!*2!}=\frac{(n-1)*n}{2}$$.

Hope it's clear.
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Re: A company that ships boxes to a total of 12 distribution [#permalink]

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20 May 2017, 20:13
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chintzzz 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 colors in a pair does not matter)

A. 4
B. 5
C. 6
D. 12
E. 24

1. Solving a simple case and then generalizing would be easy for this problem.
2. Take 2 colors Red and Blue. These two can be used in the following ways R, B, RB. i.e, 2+2C2. It can represent only 3 centers
3. Take 3 colors R, B, G. These can represent 3 +3c2=6 centers
4. Four colors can represent 4+4C2= 10 centers
5 colors can represent 5+5C2=15 centers

So we see a minimum of 5 colors are needed
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Re: A company that ships boxes to a total of 12 distribution [#permalink]

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21 May 2017, 08:44
see if we have 4 different colors then we have 4 unique single identity and 4*3/1*2=6 unique identity with pairs so slightly more than 4 will be the answer that is 5

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Re: A company that ships boxes to a total of 12 distribution [#permalink]

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02 Jun 2017, 11:11
chintzzz 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 colors in a pair does not matter)

A. 4
B. 5
C. 6
D. 12
E. 24

For this problem, you can simply list out the possibilities. Since it's a min/max problem, starting with A is best. Repetition of colors is not allowed.

A) n = 4

Let ABCD represent four colors.

ABCD = 4 centers covered
AB AC AD = 3 more centers
BC BD = 2 more centers
CD = 1 more center

The total here is 11. Since we are close to 12, an increase in one color should be more than enough. B is the answer.

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Re: A company that ships boxes to a total of 12 distribution [#permalink]

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02 Jun 2017, 12:11
Solve this using algebra

nC2 + nC1 >= 12

n*(n-1)/2 + n >= 12

Testing the answer options , 5 is the minimum value which satisfies the condition.

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Re: A company that ships boxes to a total of 12 distribution   [#permalink] 02 Jun 2017, 12:11
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