Suppose we want to choose 3 colors, without replacement, from the 4 colors red, blue, green, and purple. \[ \begin{array}{l}\text { (a) How many ways can this be done, if the order of the choices is taken into } \\ \text { consideration? } \\ \text { (b) How many ways can this be done, if the order of the choices is not taken into } \\ \text { consideration? }\end{array} \] (a)

To solve the problem, we will address both parts (a) and (b) separately. ### (a) Choosing 3 colors with order considered When the order of the choices is taken into consideration, we are looking for permutations of 3 colors chosen from 4 colors. The formula for permutations of \( r \) items from \( n \) items is given by: \[ P(n, r) = \frac{n!}{(n - r)!} \] In this case, \( n = 4 \) (the total number of colors) and \( r = 3 \) (the number of colors we want to choose). Thus, we have: \[ P(4, 3) = \frac{4!}{(4 - 3)!} = \frac{4!}{1!} = \frac{4 \times 3 \times 2 \times 1}{1} = 4 \times 3 \times 2 = 24 \] So, the number of ways to choose 3 colors with order considered is **24**. ### (b) Choosing 3 colors without order considered When the order of the choices is not taken into consideration, we are looking for combinations of 3 colors chosen from 4 colors. The formula for combinations of \( r \) items from \( n \) items is given by: \[ C(n, r) = \frac{n!}{r!(n - r)!} \] In this case, \( n = 4 \) and \( r = 3 \). Thus, we have: \[ C(4, 3) = \frac{4!}{3!(4 - 3)!} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \cdot 1} = \frac{4}{1} = 4 \] So, the number of ways to choose 3 colors without order considered is **4**. ### Summary of Results - (a) The number of ways to choose 3 colors with order considered: **24** - (b) The number of ways to choose 3 colors without order considered: **4**