• 5 Day FREE Trial
Study Smarter, Not Harder

Available with Beat the GMAT members only code

• Get 300+ Practice Questions

Available with Beat the GMAT members only code

• Award-winning private GMAT tutoring
Register now and save up to \$200

Available with Beat the GMAT members only code

• Free Practice Test & Review
How would you score if you took the GMAT

Available with Beat the GMAT members only code

• Magoosh
Study with Magoosh GMAT prep

Available with Beat the GMAT members only code

• Free Trial & Practice Exam
BEAT THE GMAT EXCLUSIVE

Available with Beat the GMAT members only code

• Most awarded test prep in the world
Now free for 30 days

Available with Beat the GMAT members only code

• 1 Hour Free
BEAT THE GMAT EXCLUSIVE

Available with Beat the GMAT members only code

• 5-Day Free Trial
5-day free, full-access trial TTP Quant

Available with Beat the GMAT members only code

• Free Veritas GMAT Class
Experience Lesson 1 Live Free

Available with Beat the GMAT members only code

## Chemicals color coding system

tagged by: LeoBen

This topic has 3 member replies
LeoBen Senior | Next Rank: 100 Posts
Joined
31 Oct 2011
Posted:
32 messages
2

#### Chemicals color coding system

Wed Jan 04, 2012 4:28 pm
In a certain lab, chemicals are identified by a color coding system. There are 20 different chemicals. Each is coded with either a single color or a unique two color combination. If the order of colors in the pairs doesnt matter, what is the miniumum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of color.

a. 5
b. 6
c. 7
d. 20
e. 40

OA: B

My answer doesnt equate OA, below is my approach (backsolving) PLEASE correct me. Thanks.

consider a. 5 as it is the smallest amongst all.

1 or 2 nos = _ + _ _

5 + 5C2 X 2! = 5 + 10 X 2! = 25 --> this is sufficient as other will give me more than this value.

Checking with the OE - the only contention is the 2! in 2 colors, my rationale being the 2 colors chosen can be arranged in two ways. For ex. we choose red & blue - these can be arranged as: RB or BR.
Analogy I thought - if we have 3 colors RAG (red, amber, green) , selecting 2 of them we can form 6 of them:
RA
AG
GR
AR
GA
RG
------------------------- Am I missing a point here?

Source: Kaplan

ArunangsuSahu Master | Next Rank: 500 Posts
Joined
31 Mar 2011
Posted:
382 messages
15
Fri Jan 06, 2012 1:01 am
Knowledge of Combination

The problem is same as Either 1 or 2 choices...

6C1+6C2=21>20

So minimum 6 Colors needed

LeoBen Senior | Next Rank: 100 Posts
Joined
31 Oct 2011
Posted:
32 messages
2
Fri Jan 06, 2012 12:02 am
Oh yes - my mistake. Ofcourse order doesnt matter, so RB or BR its just the same as per the stem i missed it I guess.

Thanks Pemdas

pemdas Legendary Member
Joined
15 Apr 2011
Posted:
1085 messages
Followed by:
21 members
158
Wed Jan 04, 2012 5:45 pm
this one is quite easy without backsolving

I'll try to explain the logic firstly

you can select two color sets without order priority among n colors -> nC2
your selection must return at most 20 - also your selection may return less than 20 (you need to trade-off here in the solution process) -> nC2 =< 20
find such value for n so that n distributed in two-color sets Plus some value of (n-value) is enough to cover up 20 chemicals

Let's see, nC2 and n=7, hmm we have nC2=21 (greater than 20) reject
nC2 and n=6, we get nC2=15, OK fine, we also have 5 colors among six individual colors which are distributed in pairs. Thus, total makes 15 two-color sets made up out of six colors + 5 solid colors out of six colors makes minimum 6 colors (revised for clarity)

b

your mistake was in permuting nP2 instead of nC2, when the problem explicitly specifies the order in pairs doesn't matter. In your way, 5P2 is 20 and you get there ans.A which is wrong BTW.
LeoBen wrote:
In a certain lab, chemicals are identified by a color coding system. There are 20 different chemicals. Each is coded with either a single color or a unique two color combination. If the order of colors in the pairs doesnt matter, what is the miniumum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of color.

a. 5
b. 6
c. 7
d. 20
e. 40

OA: B

My answer doesnt equate OA, below is my approach (backsolving) PLEASE correct me. Thanks.

consider a. 5 as it is the smallest amongst all.

1 or 2 nos = _ + _ _

5 + 5C2 X 2! = 5 + 10 X 2! = 25 --> this is sufficient as other will give me more than this value.

Checking with the OE - the only contention is the 2! in 2 colors, my rationale being the 2 colors chosen can be arranged in two ways. For ex. we choose red & blue - these can be arranged as: RB or BR.
Analogy I thought - if we have 3 colors RAG (red, amber, green) , selecting 2 of them we can form 6 of them:
RA
AG
GR
AR
GA
RG
------------------------- Am I missing a point here?

Source: Kaplan

_________________
Success doesn't come overnight!

### Best Conversation Starters

1 lheiannie07 108 topics
2 Roland2rule 63 topics
3 ardz24 63 topics
4 LUANDATO 50 topics
5 AAPL 42 topics
See More Top Beat The GMAT Members...

### Most Active Experts

1 GMATGuruNY

The Princeton Review Teacher

152 posts
2 Jeff@TargetTestPrep

Target Test Prep

106 posts
3 Rich.C@EMPOWERgma...

EMPOWERgmat

104 posts
4 Scott@TargetTestPrep

Target Test Prep

96 posts
5 Max@Math Revolution

Math Revolution

87 posts
See More Top Beat The GMAT Experts