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Use a tree diagram to find the number of ways 3 letters can be chosen from the set \( \{P, Q, R\} \) if order is important and a. if repetition is allowed; b. if no repeats are allowed; c. Find the number of combinations of 3 elements taken 3 at a time. Does this answer differ from that in part (a) or (b)? a. If repetition is allowed, how many ways can 3 letters are chosen from the set \( \{P, Q, R\} \) if order is important? b. If no repeats are allowed, how many ways can 3 letters are chosen from the set \( \{P, Q, R\} \) if order is important? c. Find the number of combinations of 3 elements taken 3 at a time.

Ask by Huff Fuentes. in the United States
Mar 09,2025

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Answer

**a. Repetition allowed (order matters):** 27 ways **b. No repeats allowed (order matters):** 6 ways **c. Combinations of 3 elements taken 3 at a time (order does not matter):** 1 way The combination answer is different from parts (a) and (b).

Solution

**a. Repetition is allowed (order matters):** Construct a tree diagram where each level represents the choice of one letter. - **Level 1:** There are 3 choices: \( P \), \( Q \), or \( R \). - **Level 2:** Since repetition is allowed, for each branch from level 1 there are again 3 choices. - **Level 3:** For each branch from level 2 there are again 3 choices. The total number of outcomes is \[ 3 \times 3 \times 3 = 27. \] **b. No repeats allowed (order matters):** Now, construct a tree diagram where a chosen letter cannot be repeated. - **Level 1:** There are 3 choices: \( P \), \( Q \), or \( R \). - **Level 2:** After choosing one letter, there remain 2 choices. - **Level 3:** Finally, only the remaining letter can be chosen, giving 1 choice. Thus, the total number of outcomes is \[ 3 \times 2 \times 1 = 6. \] **c. Combinations of 3 elements taken 3 at a time (order does not matter):** When order is not important, we use combinations. The number of combinations is given by \[ \binom{3}{3} = \frac{3!}{3!(3-3)!} = \frac{3!}{3! \cdot 0!} = 1, \] since \(0! = 1\). **Comparison:** - Part (a) (with repetition & order matters): 27 outcomes. - Part (b) (without repetition & order matters): 6 outcomes. - Part (c) (order does not matter): 1 outcome. Thus, the answer obtained in part (c) is different from both parts (a) and (b).

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The Deep Dive

To tackle your queries involving the set \( \{P, Q, R\} \), let's break it down step by step! **a. If repetition is allowed:** Since order is important and we can repeat letters, each of the three positions can be filled with any of the three letters. Therefore, the total number of ways to choose 3 letters is given by: \[ 3 \times 3 \times 3 = 3^3 = 27 \] **b. If no repeats are allowed:** Here, order is still important, but we can't use the same letter more than once. For the first letter, we have 3 choices (P, Q, or R). For the second letter, since one letter has been used, we have 2 choices left. Finally, we have only 1 choice for the last letter. Thus, the total number of ways to arrange these letters is: \[ 3 \times 2 \times 1 = 6 \] **c. Combinations of 3 elements taken 3 at a time:** When selecting combinations, order does not matter, and in this case, we are choosing all 3 letters from the set \( \{P, Q, R\} \). There is only one way to choose all three letters: {P, Q, R}. Thus, the number of combinations is: \[ 1 \] As for your final thought: the answers to parts (a) and (b) are indeed different from the result in part (c). In part (c), we considered combinations where order does not matter and duplicates are not allowed, resulting in a single selection, whereas parts (a) and (b) focused on permutations where the arrangement and repetition rules changed the outcomes significantly.

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