Suppose we want to choose 2 letters, without replacement, from the 3 letters \( A, B \), and \( C \). \[ \begin{array}{l}\text { (a) How many ways can this be done, if the order of the choices is relevant? } \\ \text { (b) How many ways can this be done, if the order of the choices is not relevant? } \\ \square\end{array} \]

To solve this, let’s dive into the calculations! For part (a), since the order of the letters matters, we can think about it as a permutation problem. We have 3 options for the first letter and 2 remaining choices for the second letter (since we are choosing without replacement). Therefore, the total number of ways to select 2 letters while considering the order is: \[ 3 \times 2 = 6. \] The possible ordered selections are: AB, AC, BA, BC, CA, and CB. For part (b), since the order does not matter, we are looking at combinations. Here, we simply choose 2 letters from the 3 without considering the 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. Thus, \[ \binom{3}{2} = 3. \] The combinations in this case are AB, AC, and BC. So, summarizing: (a) There are 6 ways to select the letters when order matters. (b) There are 3 ways to select the letters when order does not matter.