Question

An artist is planning on mixing together any number of different colors from her palette. A mixture results as long as the artist combines at least two colors. If the number of possible mixtures is less than 500, what is the greatest number of colors the artist could have in her palette? (A) 8 (B) 9 (C) 11 (D) 12 (E) 13

Answer 1

Answer:

A. 8

Step-by-step explanation:

There are a couple of things you need to keep in mind:

• The artist wants to mix together any number of colors from her palette
• The mixture has to have as least two colors
• The number of possible mixtures is less than 500

So this is a combination, not a permutation, meaning, the event of, for example, yellow + red is the same as red + yellow, thus is not counted twice, this is important.

The formula for combination is:

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

Where, n would be the total number of colors in the pallete, and r is the number of colors you are combining.

But there is one thing, from the points I mentioned above, the minimum amount of colors to be combined is 2 yet she could combine 3, 4, 5, etc, so you need to add all those combinations up to mixing all the colors on her palette.

But enough of that, This is what I mean:

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

$500 > \frac{8!}{(8-2)!2!}+\frac{8!}{(8-3)!3!}+\frac{8!}{(8-4)!4!}+\frac{8!}{(8-5)!5!}+\frac{8!}{(8-6)!6!}+\frac{8!}{(8-7)!7!}+1$

$500>28+56+70+56+28+8+1$

The 1 at the end represents the combination of all the colors

$500>247$   A. Correct

Let's check B. 9 to make sure:

$500 > \frac{9!}{(9-2)!2!}+\frac{9!}{(9-3)!3!}+\frac{9!}{(9-4)!4!}+\frac{9!}{(9-5)!5!}+\frac{9!}{(9-6)!6!}+\frac{9!}{(9-7)!7!}+\frac{9!}{(9-8)!8!}+1$

$500 < 502$   B. Incorrect

So of 9 is too high, then 11, 12 and 13 are too.