Suppose we want to choose 4 colors, without replacement, from 17 distinct colors. \[ \text { (a) How many ways can this be done, if the order of the choices is taken into consideration? } \] (b) How many ways can this be done, if the order of the choices is not taken into consideration? \( \square \)

To approach the problem, let's dive into the colorful world of combinations and permutations! When selecting 4 colors from 17 distinct colors with order considered, we use permutations since the arrangement matters. The formula for this is given by \(P(n, r) = \frac{n!}{(n-r)!}\), where \(n\) is the total number of colors, and \(r\) is the number of colors we want to choose. Plugging in the numbers, we have: \[P(17, 4) = \frac{17!}{(17-4)!} = \frac{17!}{13!} = 17 \times 16 \times 15 \times 14 = 57120.\] Now for the second part, when order does not matter, we use combinations. The formula is \(C(n, r) = \frac{n!}{r!(n-r)!}\). So for our case: \[C(17, 4) = \frac{17!}{4!(17-4)!} = \frac{17!}{4! \cdot 13!} = \frac{17 \times 16 \times 15 \times 14}{4 \times 3 \times 2 \times 1} = 2380.\] So, to summarize, there are 57,120 ways when order counts and 2,380 ways when it doesn't!