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?

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 +56945517215

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.