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

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Math Expert
Joined: 02 Sep 2009
Posts: 43831
A company that ships boxes to a total of 12 distribution cen [#permalink]

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06 Mar 2014, 01:57
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Difficulty:

35% (medium)

Question Stats:

67% (01:11) correct 33% (01:22) wrong based on 607 sessions

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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

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

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06 Mar 2014, 01:58
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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) $$n\geq{5}$$ --> $$n_{min}=5$$.

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

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06 Mar 2014, 02:18
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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!

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

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06 Mar 2014, 03:16
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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 the colors in a pair does not matter.)

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

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........
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Joined: 26 Feb 2012
Posts: 14
GMAT 3: 640 Q49 V29
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

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06 Mar 2014, 03:19
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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,

$$n2+n = 29 >= 24$$

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

Math Expert
Joined: 02 Sep 2009
Posts: 43831
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

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08 Mar 2014, 11:46
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Manager
Joined: 05 Jul 2015
Posts: 107
GMAT 1: 600 Q33 V40
GPA: 3.3
Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

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22 Nov 2015, 10:48
1
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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.
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Re: A company that ships boxes to a total of 12 distribution cen [#permalink]

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25 Nov 2015, 14:31
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).

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

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08 Jan 2018, 09:20
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Re: A company that ships boxes to a total of 12 distribution cen   [#permalink] 08 Jan 2018, 09:20
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