Problem Solving

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

[spoiler]OA: B[/spoiler]

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

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 [spoiler](revised for clarity)[/spoiler]

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

[spoiler]OA: B[/spoiler]

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

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

Knowledge of Combination

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

6C1+6C2=21>20

So minimum 6 Colors needed

(B) is the answer