In a certain laboratory, chemicals are identified by a color-coding sy

Let’s analyze the answer choices.

If there are 5 different colors, the number of codes that can be formed is 5C1 + 5C2 = 5 + 10 = 15, which is not sufficient for the 20 different chemicals.

If there are 6 different colors, the number of codes that can be formed is 6C1 + 6C2 = 6 + 15 = 21, which IS sufficient for the 20 different chemicals. Therefore, we need at least 6 different colors.

Answer: B
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Shouldn't the approach to solve this question be the following:

Let the number of colours be 'n', and we have n + n(n-1)/2 >= 20

We get 2n + n^2 - n = 40, simplifying further we get n^2 + n = 40

Now, putting in the values (since we only require the minimum no. of colors), we see that n cannot be equal to 5 and 6 is the first (and the minimum) value that satisfies the equation.

Hence, option (B)

6c1 +6c2 : 21

IMO 6 should be correct.. IMO b

Let say there are 2 colors : A & B. All possible combination would be A , B , AB.
If there are 3 : A, B & C. Combinations would be A , B , C , AB, AC , BC.

n(n+1)/ 2 > 20
n(n+1) > 40

Lowest value which fits is 6.

Ans B

In a certain laboratory, chemicals are identified by a color-coding system. There are 20 different chemicals. Each one is coded with either a single color or a unique two-color pair. If the order of colors in the pairs doesn't matter, what is the minimum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of colors?

CONSIDER ANSWER CHOICES

5 colours

A,B,C,D,E
AB, AC, AD, AE
BC, BD, BE
CD, CE
DE
falls short of 20.

6 colours

all the above + a new colour F
F, FA, FB, FC,FD, FE
totals 21.

B is the correct option.

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