Counting problem: Color code

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psm12se: Master | Next Rank: 500 Posts; Posts: 109; Joined: Sun Oct 21, 2012 5:10 amFollowed by: 1 member

by psm12se » Sat Dec 07, 2013 4:24 am

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

OA: E

I solved this problem correctly but using the counting method and was wondering if the client count was more, I mean 50 or 80. What should the method of solving be in this case.

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Source: Beat The GMAT — Problem Solving | Sat Dec 07, 2013 4:24 am

Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members; Location: Chennai, India; Thanked: 83 times

by Uva@90 » Sat Dec 07, 2013 5:22 am

psm12se 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

OA: E

Hi Psm12se,
Since it has more single digit values we can count the number of possibles.
Take the minimum answer choice given in the question, i.e 5
let color's be A,B,C,D,E
So,
A,B,C,D,E,AB,AC,AD,AE,BC,BD,BE,CD,CE,DE.
Which is more than 12. hence 5 is sufficient.
Answer is E

Regards,
Uva.

Known is a drop Unknown is an Ocean

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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Sat Dec 07, 2013 5:23 am

If we use 4 colors ==> 4 + 4c2 => 4 + 4x3/2 => 4 + 6 = 10 NO

If we use 5 colors ==> 5 + 5C2 => 5 + 5x4/2 => 5 + 10 = 15 YES

Answer [spoiler]{E}[/spoiler]

R A H U L

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Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members; Location: Chennai, India; Thanked: 83 times

by Uva@90 » Sat Dec 07, 2013 5:26 am

I solved this problem correctly but using the counting method and was wondering if the client count was more, I mean 50 or 80. What should the method of solving be in this case.

It can be found using,
nC2 + n
in this case,
Let take n = 4
4C2+4 = 10
n=5
5C2+5 = 15 , which is more than 12.
hence N = 5
Answer is E

Regards,
Uva.[/quote]

Known is a drop Unknown is an Ocean

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GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sat Dec 07, 2013 7:25 am

I thought I'd mention that this question is based on a question from the Official Guide (OG): https://www.beatthegmat.com/distribution ... tml#678097

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GMAT/MBA Expert

[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members; Location: Palo Alto, CA; Thanked: 2867 times; GMAT Score:800

by [email protected] » Sat Dec 07, 2013 3:19 pm

Hi psm12se,

You'll notice in the wording of the prompt that a 2-color code can be used and that the order of the colors DOES NOT matter. This wording implies that we'll be dealing with "combinations" of colors, which means that we could use the "combination formula" to answer this question.

Usually, when the GMAT offers this type of question during the exam, the answers are smaller, and more manageable, than you might expect. That's because the Quant section is NOT a "math test", so most questions can be solved in ways that are not, strictly speaking, just calculations. Keep that in mind as you continue to study.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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psm12se: Master | Next Rank: 500 Posts; Posts: 109; Joined: Sun Oct 21, 2012 5:10 amFollowed by: 1 member

by psm12se » Mon Dec 09, 2013 4:22 am

Thanks Rich for the insight.

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