What is the minimum number of colours needed for coding 12 objects, if each may be marked with either one or two colours?

Question

I have a word problem here which is a kind of high level to me

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A company that ships boxes to a total of (12) distribution centres uses colour coding to identify each centre. If either a single colour or a pair of two different colours is chosen to represent each centre and if each centre is uniquely represented by that choice of one or two colours, what is the minimum number of colours needed for the coding? (Assume that the order of the colours in a pair does not matter).

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I have typed the AS IS from the text book.

If there are any practice questions like this please specify the link, I want to solve these problems more..to develop myself..

Thank you all

Answer 1

With $n$ colors you can make $n$ single color labels and ${n\choose 2}$ two-color labels, giving a total of $$N_n=n+{n(n-1)\over2}={n(n+1)\over2}$$ possibilities. Since $N_4=10<12<15=N_5$ you need $5$ colors for the task.

Answer 2

You'll need 5 colours. With n colours, you can make $\sum\limits_{i=1}^n i$ combinations

Answer 3

I would write this as a comment on @Mirac7's answer, but I don't have enough reputation to do that yet. Here's my take on a combinatorial reasoning for why $n$ colors correspond to $\sum_{i=1}^n i$ combinations.

With 1 color (red), you can identify 1 thing.
With a second color (blue), you can identify 2 more things: red/blue and blue.
With a third color (violet), you can identify 3 additional things: red/violet, blue/violet, and violet.
With a fourth color (black), you can identify 4 additional things: red/black, blue/black, violet/black, and black.

In general, when you're given the $i$th color, you can identify $i$ additional things: you get 1 identifier by using the new color by itself, and you get $i-1$ by combining the new color with each of the $i-1$ colors you already had.