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by jc114 » Fri Apr 13, 2007 7:42 am
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 2 colors, what is the minimum # 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
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by jaspetrovic » Fri Apr 13, 2007 8:15 am
Hi there

Tere are 2 ways to solve this question.

As I first saw this q. I knew that it tests the use of combinations, but I used a little bit different app. Itlooked faster to me.
I looked at the ansers and started to backsolve. I marked colors w/numbers: 1, 2, 3, 4,...
So, in A - we have 4 combinations w/one color and 6 combinations w/2 colors(12, 13, 14, 23, 24, 34) => 4 + 6=10<12 NOT enough

B - 5 comb. w/ one color and 10 comb. w/2 colors (12, 13, 14, 15, 23, 24, 25, 34, 35, 45) => 5+10=15>12 MORE than enough


Later I saw the 2nd app. through use of combination formula.
Ok, so in A we have 4 colors so we need 8 comb. of two colors (12-4). 4!/2!(4-2)!=6<8. We need more colors. If we use 5 color we would need (12-5) 7 comb. of 2 colors => 5!/2!(5-2)!=10>7

So, B is the answer
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by Cybermusings » Sat Apr 14, 2007 2:58 am
Is the answer 5?
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by Scott@TargetTestPrep » Mon May 20, 2019 6:08 pm
jc114 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 2 colors, what is the minimum # 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
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.

Answer: B

Scott Woodbury-Stewart
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by Brent@GMATPrepNow » Tue May 21, 2019 6:31 am
jc114 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 2 colors, what is the minimum # 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
We need to be able to create AT LEAST 12 codes (to represent the 12 countries).

Let's test the options.
Can we get 12 or more color codes with 4 colors?
Let's see . . .

1-color codes = 4 (since there are 4 colors)
2-color 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, here's a video on calculating combinations (like 4C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Can we get 12 or more color codes with 5 colors?
1-color codes = 5 (since there are 5 colors)
2-color 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 5 (B)

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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