Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
John has 12 clients and he wants to use color coding to iden
[#permalink]
Show Tags
03 Jan 2011, 18:02
1
5
00:00
A
B
C
D
E
Difficulty:
25% (medium)
Question Stats:
72% (00:57) correct 28% (01:07) wrong based on 270 sessions
HideShow timer Statistics
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.
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
Let # of colors needed be \(n\), then it must be true that \(n+C^2_n\geq{12}\) (\(C^2_n\) - # of ways to choose the pair of different colors from \(n\) colors when order doesn't matter) --> \(n+\frac{n(n-1)}{2}\geq{12}\) --> \(2n+n(n-1)\geq{24}\) --> \(n(n+1)\geq{24}\) --> as \(n\) is an integer (it represents # of colors) \(n\geq{5}\) --> \(n_{min}=5\).
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)
John has 12 clients and he wants to use color coding to iden
[#permalink]
Show Tags
01 Aug 2015, 09:00
2
SQUINGEL wrote:
Is there any other way to solve this question?
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.
Re: John has 12 clients and he wants to use color coding to iden
[#permalink]
Show Tags
05 Aug 2015, 01:34
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
John has 12 clients and he wants to use color coding to iden
[#permalink]
Show Tags
05 Aug 2017, 12:27
rtaha2412 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
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
_________________
In the depths of winter, I finally learned that within me there lay an invincible summer. -- Albert Camus, "Return to Tipasa"
Re: John has 12 clients and he wants to use color coding to iden
[#permalink]
Show Tags
07 Jul 2018, 07:45
Top Contributor
rtaha2412 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
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)
_________________
Brent Hanneson – GMATPrepNow.com Sign up for our freeQuestion of the Dayemails
gmatclubot
Re: John has 12 clients and he wants to use color coding to iden &nbs
[#permalink]
07 Jul 2018, 07:45
close
72% (00:57) correct 
