Suppose we want to choose 2 colors, without replacement, from the 4 colors red, blue, green, and purple. (a) How many ways can this be done, if the order of the choices matters? (b) How many ways can this be done, if the order of the choices does not matter? \( \square \)

Let's solve each part step by step. ### (a) Order Matters (Permutations) When the order of selection matters, we're dealing with **permutations**. - **First Choice:** There are 4 possible colors. - **Second Choice:** Since we're not replacing the first color, there are 3 remaining colors. **Total number of ways:** \[ 4 \times 3 = 12 \] **Answer:** There are **12** ways when the order matters. ### (b) Order Does Not Matter (Combinations) When the order of selection does not matter, we're dealing with **combinations**. The number of ways to choose 2 colors out of 4 without considering the order is given by the combination formula: \[ \binom{4}{2} = \frac{4!}{2! \times (4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] **Answer:** There are **6** ways when the order does not matter. \(\square\)