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? (a)
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If the order of the choices matters, we can treat this as a permutation problem. For the first color, you have 4 options (red, blue, green, purple). After selecting the first color, only 3 colors remain for the second choice. Thus, the total number of ways is calculated as \(4 \times 3 = 12\). If the order of the choices does not matter, we can use combinations. Here, you want to choose 2 colors out of 4 without regard to order. The formula for combinations is \(\binom{n}{r}\), where \(n\) is the total number of items to choose from and \(r\) is the number of items to choose. So, we calculate \(\binom{4}{2} = \frac{4!}{2!(4-2)!} = 6\). In summary: (a) 12 ways (order matters) (b) 6 ways (order does not matter)
