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# M04-29

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Math Expert
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Re: M04-29 [#permalink]
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Bunuel wrote:
John has 12 clients and he wants to use color coding to identify each client. If either a single color or a pair of two different colors can represent a client code, what is the minimum number of colors needed for the coding? Assume that changing the color order within a pair does not produce different codes.

A. 24
B. 12
C. 7
D. 6
E. 5

Let minimum n colors are required
We can write it as

nC1 + nC2 = > 12

Then try to put values from middle
7C1 + 7C2 = > 7+21 = 29

lets try lower value, if we can get value near to 12

Lets take n=5
5C1 + 5C2 = 5 + 10 = 15

This matches our requirement. We do not have smaller value than 5. So 5 is the answer
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Re: M04-29 [#permalink]
I wonder of this is a real GMAT question?

This question is asking for the min # of colors than can make >= 12 color combinations (single color + twin color).

So, lets start with 2 color code set:
If we have 4 colors: the total 2 color combo's will be 4X3=12. But its said that order doesn't matter. In other words red+green=green+red. So, 12/2 = 6 two color codes. What additional # of codes do we get by using the same 4 colors individually...it's 4. So total color code combinations = 6+4 =10......But we need 12.

So, doing the same analysis with with 5 colors, we get....(two color codes)[(5x4)/2] + (1 color codes)[5] = 10+5=15......Answer=5
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Re: M04-29 [#permalink]
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rohitd80 wrote:
I wonder of this is a real GMAT question?

This question is asking for the min # of colors than can make >= 12 color combinations (single color + twin color).

So, lets start with 2 color code set:
If we have 4 colors: the total 2 color combo's will be 4X3=12. But its said that order doesn't matter. In other words red+green=green+red. So, 12/2 = 6 two color codes. What additional # of codes do we get by using the same 4 colors individually...it's 4. So total color code combinations = 6+4 =10......But we need 12.

So, doing the same analysis with with 5 colors, we get....(two color codes)[(5x4)/2] + (1 color codes)[5] = 10+5=15......Answer=5

Why would you start from 2 when the minimum number in options is 5!

If 5 Simply make combinations as it wont take time to calculate (I struggle with formulas, my bad)
A B C D E
AB AC AD AE
BC BD BE
We are done.
5 is the answer.
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Re: M04-29 [#permalink]
Thanks Bunuel. I think I really need to go through my fundamentals on this again.
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Re: M04-29 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M04-29 [#permalink]
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start from E.

if we have 5 colors and we want to know how many 2 colors combination can give us, then 5C2 = 10. ok so best case scenario in this case would be if we have 10 two color code, hold on, we still have 5 individual colors that we can assign to the remaining 2 clients. so, we don't need to continue, 5 is the answer.
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Re: M04-29 [#permalink]
Expert Reply
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M04-29 [#permalink]
Bunuel wrote:
Official Solution:

John has 12 clients, and he wants to use color coding to identify each of them. He can use either a single color or a pair of two different colors to represent a client code. Assuming that switching the order of colors within a pair does not create a different code, what is the minimum number of colors needed for this coding scheme?

A. 24
B. 12
C. 7
D. 6
E. 5

Combination approach:

Let the number of colors needed be $$n$$. Then, ensuring that the total number of unique color codes (single color codes + two-color codes) is sufficient for all 12 clients, the inequality $$n + C^2_n \ge 12$$ must hold.

$$n+\frac{n(n-1)}{2} \ge 12$$;

$$2n+n(n-1) \ge 24$$;

$$n(n+1) \ge 24$$. As $$n$$ is an integer (it represents the number of colors), then $$n \ge 5$$, so $$n_{\text{min=5$$.

Trial and error approach:

If the minimum number of colors needed is 4, then there are 4 single color codes possible PLUS $$C^2_4=6$$ two-color codes. Total: $$4+6=10 < 12$$. Not enough for 12 codes;

If the minimum number of colors needed is 5, then there are 5 single color codes possible PLUS $$C^2_5=10$$ two-color codes. Total: $$5+10=15 > 12$$. More than enough for 12 codes.

As the least answer choice is 5, if you tried it first, you would arrive at the correct answer immediately.

Answer: E

hey Bunuel! there is quite some nuance in this question which can make one easily overlook and land up in the wrong answer, which are close choices present in the options as well. so i think it falls more in 650-700 level question. doesn't it?
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Re: M04-29 [#permalink]
Expert Reply
ganeshABC wrote:
Bunuel wrote:
Official Solution:

John has 12 clients, and he wants to use color coding to identify each of them. He can use either a single color or a pair of two different colors to represent a client code. Assuming that switching the order of colors within a pair does not create a different code, what is the minimum number of colors needed for this coding scheme?

A. 24
B. 12
C. 7
D. 6
E. 5

Combination approach:

Let the number of colors needed be $$n$$. Then, ensuring that the total number of unique color codes (single color codes + two-color codes) is sufficient for all 12 clients, the inequality $$n + C^2_n \ge 12$$ must hold.

$$n+\frac{n(n-1)}{2} \ge 12$$;

$$2n+n(n-1) \ge 24$$;

$$n(n+1) \ge 24$$. As $$n$$ is an integer (it represents the number of colors), then $$n \ge 5$$, so $$n_{\text{min=5$$.

Trial and error approach:

If the minimum number of colors needed is 4, then there are 4 single color codes possible PLUS $$C^2_4=6$$ two-color codes. Total: $$4+6=10 < 12$$. Not enough for 12 codes;

If the minimum number of colors needed is 5, then there are 5 single color codes possible PLUS $$C^2_5=10$$ two-color codes. Total: $$5+10=15 > 12$$. More than enough for 12 codes.

As the least answer choice is 5, if you tried it first, you would arrive at the correct answer immediately.

Answer: E

hey Bunuel! there is quite some nuance in this question which can make one easily overlook and land up in the wrong answer, which are close choices present in the options as well. so i think it falls more in 650-700 level question. doesn't it?

We do not assign the difficulty level manually. The difficulty level of a question on the site is determined automatically based on various parameters collected from users' attempts, such as the percentage of correct answers and the time taken to answer the question. You can find the difficulty level of a question and its related statistics in the first post.
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Re: M04-29 [#permalink]
oh okay. thanks for the clarification!

Bunuel wrote:
ganeshABC wrote:
Bunuel wrote:
Official Solution:

John has 12 clients, and he wants to use color coding to identify each of them. He can use either a single color or a pair of two different colors to represent a client code. Assuming that switching the order of colors within a pair does not create a different code, what is the minimum number of colors needed for this coding scheme?

A. 24
B. 12
C. 7
D. 6
E. 5

Combination approach:

Let the number of colors needed be $$n$$. Then, ensuring that the total number of unique color codes (single color codes + two-color codes) is sufficient for all 12 clients, the inequality $$n + C^2_n \ge 12$$ must hold.

$$n+\frac{n(n-1)}{2} \ge 12$$;

$$2n+n(n-1) \ge 24$$;

$$n(n+1) \ge 24$$. As $$n$$ is an integer (it represents the number of colors), then $$n \ge 5$$, so $$n_{\text{min=5$$.

Trial and error approach:

If the minimum number of colors needed is 4, then there are 4 single color codes possible PLUS $$C^2_4=6$$ two-color codes. Total: $$4+6=10 < 12$$. Not enough for 12 codes;

If the minimum number of colors needed is 5, then there are 5 single color codes possible PLUS $$C^2_5=10$$ two-color codes. Total: $$5+10=15 > 12$$. More than enough for 12 codes.

As the least answer choice is 5, if you tried it first, you would arrive at the correct answer immediately.

Answer: E

hey Bunuel! there is quite some nuance in this question which can make one easily overlook and land up in the wrong answer, which are close choices present in the options as well. so i think it falls more in 650-700 level question. doesn't it?

We do not assign the difficulty level manually. The difficulty level of a question on the site is determined automatically based on various parameters collected from users' attempts, such as the percentage of correct answers and the time taken to answer the question. You can find the difficulty level of a question and its related statistics in the first post.
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Re M04-29 [#permalink]
Doesn't - "He can use either a single color or a pair of two different colors to represent a client code." mean that either 1 color - i.e. 12 colors or nC2=12...not a combination of both?
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Re: M04-29 [#permalink]
Expert Reply
Ishita2000 wrote:
Doesn't - "He can use either a single color or a pair of two different colors to represent a client code." mean that either 1 color - i.e. 12 colors or nC2=12...not a combination of both?

No, that means the code can consist of either one color OR two colors.
Re: M04-29 [#permalink]
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