Question

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

• A. 7
• B. 6
• C. 5
• D. 20

Answer 1

Answer: (D)

20

Answer 2

To minimize the number of different colors needed to code all 20 chemicals with either a single color or a unique pair of colors, we should first code as many chemicals as possible with a single color because that would require the fewest colors.
First, we can code 10 chemicals with 10 unique single colors. That accounts for half of the chemicals.
Now, we need to code the remaining 10 chemicals using unique two-color pairs. To do this, we can use combinations of the colors we have already used.
The total number of unique two-color pairs that can be formed from 10 colors is given by the combination formula C(n, k), where n is the total number of colors (10) and k is the number of colors chosen at a time
$(2):c(10,2)=\frac{10!}{[(2!(10-2)!)]}=45$
So, we can code the remaining 10 chemicals with 10 unique two-color pairs.
Therefore, the minimum number of different colors needed to code all 20 chemicals is 10 (for the single colors) + 10 (for the two-color pairs) = 20.
The correct option is (D): 20