Use a tree diagram to find the number of ways 2 letters can be chosen from the set \( \{P, Q, R, S\} \) if order is important and a. if repetition is allowed; b. if no repeats are allowed; c. Find the number of combinations of 4 elements taken 2 at a time. Does this answer differ from that in part (a) or (b)? a. If repetition is allowed, how many ways can 2 letters are chosen from the set \( \{P, Q, R, S\} \) if order is important? b. If no repeats are allowed, how many ways can 2 letters are chosen from the set \( \{P, Q, R, S\} \) if order

**Part (a): Repetition Allowed (Order is Important)** Since there are 4 letters in the set \( \{P, Q, R, S\} \), when choosing 2 letters with repetition allowed, the first letter has 4 choices and the second letter also has 4 choices. Thus, the total number of ways is \[ 4 \times 4 = 16. \] A tree diagram for this situation can be envisioned as follows: - **First Letter: P** - Second Letter: P, Q, R, S - **First Letter: Q** - Second Letter: P, Q, R, S - **First Letter: R** - Second Letter: P, Q, R, S - **First Letter: S** - Second Letter: P, Q, R, S Each branch from the first letter leads to 4 branches for the second letter, totaling \( 4 \times 4 = 16 \) outcomes. --- **Part (b): No Repetition Allowed (Order is Important)** If repetition is not allowed, after choosing the first letter, only 3 letters remain for the second selection. Therefore, the total number of ways is \[ 4 \times 3 = 12. \] A tree diagram for this case is: - **First Letter: P** - Second Letter: Q, R, S - **First Letter: Q** - Second Letter: P, R, S - **First Letter: R** - Second Letter: P, Q, S - **First Letter: S** - Second Letter: P, Q, R This gives a total of \( 4 \times 3 = 12 \) outcomes. --- **Part (c): Combinations of 4 Elements Taken 2 at a Time** When the order is not important (combinations), we use the binomial coefficient \[ \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6. \] Thus, there are 6 ways to choose 2 letters from the set if order does not matter. **Comparing the Answers:** - The answer in part (a) (order matters, repetition allowed) is \(16\). - The answer in part (b) (order matters, no repetition) is \(12\). - The answer in part (c) (order does not matter, i.e., combinations) is \(6\). Each count differs because they are based on different rules: - Part (a) allows repeats and considers order, leading to more outcomes. - Part (b) does not allow repeats and still considers order. - Part (c) does not consider order, so many of the outcomes in part (b) are counted as one combination.