John has 12 clients and he wants to use color coding to iden

Question

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

Bunuel · Accepted Answer

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_{	ext{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

Hope it's clear.

ENGRTOMBA2018 · Answer

Well 'counting' is the only method applicable for this question.

As you are being asked to find the 'minimum' value, use the options to guide you.

Options A and B are out because of obvious reasons.

Start with E, Lets say you have 5 colors. Out of these 5 colors, look at how many 2 color combinations you can create = 5C2 = 10 and remaining 2 can be single colors. So there you go, you have your answer. An answer that will give you possible number of combinations  \geq  12 will be the answer as you need to cover all 12 clients  uniquely .

If lets say the total number of clients would have been = 23, then with n = 5, you could at most have = 5C2 + 5 = 15 (<23) different ways, with n =6 you could have 6C2 + 6 = 21 different ways (<23). Thus n = 7 would have been the minimum number of colors.

Hope this helps.

Burnkeal · Answer

I started with the smallest answer of all and worked my way up or so I thought ;-)

5 different colours = 5 single colours = 5 different clients There are 7 clients to go The use of the factorial is the easiest way to solve this kind of problem for me. A B C D E (different colour) C C N N N (2 colour coding + 3 colour non chosen) 5!/2!3!= 10 15 codings can be done with 5 different colours Ans E

Ralphcuisak · Answer

Bunuel,how did you calculate n(n+1)>=24. Is there any other way than putting in values and validating the equation to get the correct answer?

Bunuel · Answer

Since n must be an integer solving by number plugging is the best approach for n(n+1)>=24.

Check  Constructing Numbers, Codes and Passwords  problems for practice.

Hope it helps.

SQUINGEL · Answer

Is there any other way to solve this question?

OptimusPrepJanielle · Answer

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

We can backsolve using the answer choices and the formula n!/(r! (n-r)!) and add the number of colors we are using (to account for the single color codes) to get the total number of possible codes. As we're asked for the minimum, we can start with 5.

n=5 (five colors) and r=2 (because we're making paired color codes)
(5!)/(2! (5-2)!) = 10 possible paired color codes
10 paired possible color codes + 5 single codes for each color used = 15 possible codes. This is the minimum.
A. 24
B.12
C. 7
D. 6
 E.  5

generis · Answer

We need only 12 codes, at least 5 of 12 are single letters. Using answer choices is really quick.

Start with (E) 5, the smallest number.
List single letters to denote colors. Then list combinations.

A | AB, AC, AD, AE
B | BC, BD, BE **
C | CD, CE
D | DE
E

**You can stop here. The total is 15, more than enough.

Answer E

BrentGMATPrepNow · Answer

We need to be able to create AT LEAST 12 codes (to represent the 12 clients).

Let's test the options,  starting with the smallest value....answer choice E

So, can we get 12 or more color codes with  5  colors?
Let's see . . . 
 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 (E)

RELATED VIDEO (calculating combinations, like 5C2, in your head)
 https://www.youtube.com/watch?v=05a0A7vjG8I

CEdward · Answer

Took me 3 minutes...too long, but didn't really use any formulas on this one.

C1 C2
C1
C2
C3 C4
C3 
C4
C1 C3
C1 C4
C2 C3
C2 C4
C5

Answer is E.

sahuanamika · Answer

As per questions we need to look for two things:
1. Single color 2.Pair of those colors
 
Here we can directly start with options:
Taking the min option n = 5

1. Single color options = 5 
2. Pairs , since its given changing the order does not change the code , so order here does not matters 
 Selecting pair(2) out of 5 = 5C2 = 10 
Total 10+5 = 15

So the ans is E

gmatchile1 · Answer

Keep in mind:

5!=5x4x3x2x1

n! = nx(n-1)x(n-2)x(n-3)x...3x2x1

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

Answer: E

EXPLANATION By Claudio Hurtado Coach GMAT QUANT

The situation presents the alternative of generating code (identification) through 1 color or two colors at a time. And it requests to generate 12 codes.

Question: What is the minimum number of colors necessary to generate those 12 codes, keeping in mind that if it occupies two colors red and green, for example, does rg and gr matter the same (represents the same code)?

The situation that gives the same rg as gr and considering that I can take 2 each time from a universe greater than two (each time I take a part of the universe and not the entire universe), we are in the presence of the combinatorial model within the topic of counting methods .

Combinatorial model: (N!)/((2!)(N-2)!)

Keep in mind:

5!=5x4x3x2x1

n! = nx(n-1)x(n-2)x(n-3)x...3x2x1

We provide:

With 1 Color

If I have only 1 color, I can only get one code

With 2 Colors

If I have two colors, I can generate 2 codes of a single color
and (2!)/((2!)(2-2)!)= 2/((2)(0!)) = 2/2 = 1 2-color code. Total 3 codes.

Note 0!=1.

With 3 Colors:

3 color code
and (3!)/((2!)(3-2)!) = (3!)/((2!)(1!)) = ((3)(2!))/((2!))= 3 two-color codes.
Total 6 Codes.

Keep in mind:

n! = (n)(n-1)(n-2)!

With 4 colors:

4 codes of one color
and (4!)/((2!)(4-2)!)=(4!)/((2!)(2!))= ((4)(3)(2!))/((2!)(2!))=((4)(3))/((2))= 6 codes of 2 colors
Total 10 Codes.

With 5 colors:

5 codes of one color
and (5!)/((2!)(5-2)!)=(5!)/((2!)(3!))= ((5)(4)(3!))/((2!)(3!))=((5)(4))/(2)=((5)(2))=10 two-code colors
Total 15 codes.

Then you need 5 colors to ensure generating the 12 codes. ANSWER E

See more GMAT QUANT exercises

BlackHat098 · Answer

hi, I believe that the key is to remember that a particular color can be used for a single code as well as a double code, eg if red is used to denote Client A, Red + Orange combo can be used to represent Client B. I was of the misconception that if Red is used for Client A, it cannot be used for any other code at all.

SwethaReddyL · Answer

hi  Bunuel    Assuming that switching the order of colors within a pair does not create a different code  - because of this statement i thought 5C2 - 10 wouldn't be right as the ques specifically mentions switching order of colour doesn't give a diff code, i picked option C - 6 individual colour and 6 pair of it.  
 what does that exact statement means?

Bunuel · Answer

That statement means a pair like red + blue is the same code as blue + red, so each two-color pair is counted only once, which is why 5C2 = 10 is correct. It does not mean you can pair a color with itself, because the question explicitly says a pair of two different colors.

John has 12 clients and he wants to use color coding to iden

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