Combinatorics Pigeonhole Problem find Max Number of Possible Different Colors such that each Sub-Group of Size 9 from 60 Contains 3 Same-Color Balls

Question

Let a box contain 60 colored balls. In each group of 9 balls, at least 3 of them are the same color.

What is the maximal number of possible colors which will allow the above condition to be true?

I have tried to think of a solution in the following way - if I find the minimal number of colors for which the condition is not true and subtract 1, I'll get the maximal number for which it is true.

The minimal number of colors for which the condition is not true will satisfy the following condition - there will be at least 1 group in which there will be balls of at least 8 different colors.

Is this the correct approach? If so, how do I continue from here?

Answer 1

Your approach may work, I haven't tried it, but I find another approach is more straightforward:

You can have $7$ colors, as you can choose $54$ of one and $6$ of the others, so every set will have $3$ of that first color. You can't have $8$ colors, as then you can pick a group with one of each of them, and it would have $8$ different, so couldn't have $3$ of one.