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
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
in the solution process) -> nC2 =< 20
