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

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

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?

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.

Is there any other way to solve this question?

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.

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

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

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)

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.

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

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

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