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

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

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06 Mar 2014, 02:57
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66% (01:33) correct 34% (01:44) wrong based on 795 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

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

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We'll be glad if you participate in development of this project:
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Re: A company that ships boxes to a total of 12 distribution cen  [#permalink]

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08 Mar 2014, 12:46
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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, 04:16
5
5
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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Re: A company that ships boxes to a total of 12 distribution cen  [#permalink]

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06 Mar 2014, 02:58
4
4
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, 03:18
4
1
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, 04:19
2
1
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

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

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22 Nov 2015, 11:48
1
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, 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).

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

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

Bunuel - why are you adding $$n$$ to $$C^2_n\geq{12}$$ and not just write $$C^2_n\geq{12}$$ without adding $$n$$ ...
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Re: A company that ships boxes to a total of 12 distribution cen  [#permalink]

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25 Feb 2018, 08:27
1
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) $$n\geq{5}$$ --> $$n_{min}=5$$.

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

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25 Feb 2018, 11:07
Nice one.
n + nC2 >= 12
n^2 + n -24>=0
Satisfies b = 5
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Re: A company that ships boxes to a total of 12 distribution cen  [#permalink]

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

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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.

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

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

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 ?

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

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23 Mar 2018, 12:21
1
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) $$n\geq{5}$$ --> $$n_{min}=5$$.

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 ?

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

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

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 ?

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

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

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

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23 Mar 2018, 14:09
1
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.
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