Came across this question in a older forum.
a company has 15 disrtibution centres and uses color coding to identify each center.either a single color or a pair of two different colors is chosen to represent each center uniquely.what is the mnimum number of colors needed for the coding and the order of the colors doesnot matter?
Dont have the OA. but would appreciate if someone expert can comment.
Answers given were either 4 colours or 5 colours.
Distribution centers
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We need to be able to create AT LEAST 12 codes (to represent the 12 countries).hjafferi wrote:A company that ships boxes to a total of 12 countries uses color coding to identify each center. If either a single color or a pair of 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 doesn't matter)
(A) 4
(B) 5
(C) 6
(D) 12
(E) 24
Let's test the options.
Can we get 12 or more color codes with 4 colors?
Let's see . . .
1color codes = 4 (since there are 4 colors)
2color codes = We need to choose 2 colors from 4. This can be accomplished in 4C2 ways (using combinations). 4C2 = 6
So, using 4 colors, the total number of color codes we can create = 4 + 6 = 10
We want to create AT LEAST 12 color codes, so we can eliminate answer choice A.
Aside: If anyone is interested, we have a free video on calculating combinations (like 4C2) in your head: https://www.gmatprepnow.com/module/gmatcounting?id=789
Can we get 12 or more color codes with 5 colors?
1color codes = 5 (since there are 5 colors)
2color codes = We need to choose 2 colors from 5. This can be accomplished in 5C2 ways (using combinations). 5C2 = 10
So, using 5 colors, the total number of color codes we can create = 5 + 10 = 15
Perfect!
The answer is [spoiler]5 = B[/spoiler]
Cheers,
Brent

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hjafferi wrote:
A company that ships boxes to a total of 12 countries uses color coding to identify each center. If either a single color or a pair of 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 doesn't matter)
(A) 4
(B) 5
(C) 6
(D) 12
(E) 24
Ipshitasaha wrote:[size=Normal]Is there any other approach to solve this question?[/size]
Yes, there is.
Say a minimum of n numbers of colors is needed.
Thus,
1. Total numbers of onecolor codes = n
2. Total numbers of twocolor codes = nC2 = n(n  1)/2; we did not apply permutation nP2 since the question states that the order of the colors in a pair doesn't matter.
Since these two together must be at least 12, we have
n + n(n  1)/2 â‰¥ 12
2n + n(n  1) â‰¥ 24
2n + n^2  n â‰¥ 24
n^2 + n  24 â‰¥ 0
The best way to solve this quadratic inequality is by pluggingin option values.
A. n = 4 => n^2 + n  24 â‰¥ 0 => 4^2 + 4  24 â‰¥ 0 => 4 â‰¤ 0. This is an invalid value.
B. n = 5 => n^2 + n  24 â‰¥ 0 => 5^2 + 5  24 â‰¥ 0 => 6 â‰¥ 0. This is a valid minimum value.
The correct answer: B
Hope this helps!
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[email protected] wrote:hjafferi wrote:
A company that ships boxes to a total of 12 countries uses color coding to identify each center. If either a single color or a pair of 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 doesn't matter)
(A) 4
(B) 5
(C) 6
(D) 12
(E) 24Ipshitasaha wrote:[size=Normal]Is there any other approach to solve this question?[/size]
Yes, there is.
Say a minimum of n numbers of colors is needed.
Thus,
1. Total numbers of onecolor codes = n
2. Total numbers of twocolor codes = nC2 = n(n  1)/2; we did not apply permutation nP2 since the question states that the order of the colors in a pair doesn't matter.
Since these two together must be at least 12, we have
n + n(n  1)/2 â‰¥ 12
2n + n(n  1) â‰¥ 24
2n + n^2  n â‰¥ 24
n^2 + n  24 â‰¥ 0
The best way to solve this quadratic inequality is by pluggingin option values.
A. n = 4 => n^2 + n  24 â‰¥ 0 => 4^2 + 4  24 â‰¥ 0 => 4 â‰¤ 0. This is an invalid value.
B. n = 5 => n^2 + n  24 â‰¥ 0 => 5^2 + 5  24 â‰¥ 0 => 6 â‰¥ 0. This is a valid minimum value.
The correct answer: B
Hope this helps!
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Jay
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If you insist for an Algebraic route from n^2 + n  24 â‰¥ 0, it is there; however, it's not suggested as this is a timekiller.
So, continuing from n^2 + n  24 â‰¥ 0
Let's make this quadratic inequality a perfect square.
n^2 + 2*n*(1/2) + [1/4  1/4]  24 â‰¥ 0
[n^2 + 2*n*(1/2) + 1/2^2]  1/4  24 â‰¥ 0
(n + 1/2)^2  (1/4 + 24) â‰¥ 0
(n + 1/2)^2 â‰¥ (1/4 + 24)
Take a square root of both the sides.
(n + 1/2) â‰¥ âˆš(~25); the closest perfect square number to (1/4 + 24) is 25.
(n + 1/2) â‰¥ ~5
n â‰¥ ~5 1/2
n â‰¥ ~4.5
The minimum integer value of n = 5.
The correct answer: B
Hope this helps!
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A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4
B. 5
C. 6
D. 7
E. 8
We can PLUG IN THE ANSWERS, which represent the least number of letters required to make 12 codes.
Start with the smallest answer choice.
A: 4
The following codes are possible for letters A, B, C and D:
A, B, C, D, AB, AC, AD, BC, BD, CD.
10 codes.
Since the number of codes is just a bit less than 12, one more letter is needed.
The correct answer is B.
B: 5
The following codes are possible for letters A, B, C, D and E:
A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE.
15 codes.
Success!
A. 4
B. 5
C. 6
D. 7
E. 8
We can PLUG IN THE ANSWERS, which represent the least number of letters required to make 12 codes.
Start with the smallest answer choice.
A: 4
The following codes are possible for letters A, B, C and D:
A, B, C, D, AB, AC, AD, BC, BD, CD.
10 codes.
Since the number of codes is just a bit less than 12, one more letter is needed.
The correct answer is B.
B: 5
The following codes are possible for letters A, B, C, D and E:
A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE.
15 codes.
Success!
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I'd also plug in the answers if you want a quick and dirty answer, but if we had more time (or the numbers were more unreasonable), we could derive a general equation.
Say that we have n letters, subject to the restrictions on coding given in the stem. Since we have n letters, we can have n single digit codes: that part's easy.
For the two digit codes, we have to put the letters in alphabetical order, so if we pick the first letter first, we have (n  1) possible codes, if we pick the second letter first, we have (n  2) possible codes, etc., for a total of (n  1) + (n  2) + ... + 1 possible codes. The sum of the integers from 1 to (n  1) is ((n  1) * n)/2, giving us that many double digit codes.
In all, then, we've got n + ((n  1) * n)/2 codes.
Applying that formula to this problem, we need
n + (n  1) * n/2 â‰¥Â 12
After hashing out the quadratic, we see that this will work for any n > â‰ˆ4.42, so n = 5 is the smallest solution.
Say that we have n letters, subject to the restrictions on coding given in the stem. Since we have n letters, we can have n single digit codes: that part's easy.
For the two digit codes, we have to put the letters in alphabetical order, so if we pick the first letter first, we have (n  1) possible codes, if we pick the second letter first, we have (n  2) possible codes, etc., for a total of (n  1) + (n  2) + ... + 1 possible codes. The sum of the integers from 1 to (n  1) is ((n  1) * n)/2, giving us that many double digit codes.
In all, then, we've got n + ((n  1) * n)/2 codes.
Applying that formula to this problem, we need
n + (n  1) * n/2 â‰¥Â 12
After hashing out the quadratic, we see that this will work for any n > â‰ˆ4.42, so n = 5 is the smallest solution.

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Whoops! Forgot to add that my general solution is for the version of the problem downthread (where the letters must be in alphabetical order). The version in which we're picking colors with no restrictions is simpler, and Brent's formula is dead on.