The problem involves determining the number of ways a painter can mix 3 different colors out of 8 available colors. Since the order of the colors being mixed does not matter, this is a combination problem.

The formula for combinations is:

$C(n, r) = \frac{n!}{r!(n-r)!}$

Here:

• $n = 8$ (the total number of colors),
• $r = 3$ (the number of colors to choose).

Substituting the values:

$C(8, 3) = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

Final Answer:

The painter can create 56 new colors.

Would you like a deeper explanation, or do you have any questions? Here are some related questions for further exploration:

1. What if the order of mixing mattered? How would the calculation differ?
2. How does the concept of permutations differ from combinations in this scenario?
3. What if the painter wanted to mix 4 colors instead of 3? How many combinations would be possible then?
4. Can you verify the result by listing all possible combinations manually?
5. What if one of the colors could not be repeated in a mix? How would this restriction affect the result?

Tip: Always double-check whether the problem requires order sensitivity (permutations) or not (combinations) to choose the right formula.