To solve this problem, we will again consider two scenarios: when the order of choices is relevant and when it is not relevant.

1. Order of the Choices is Relevant:

• Since the order matters, we use permutations.
• We are choosing 3 letters from a set of 4 letters (A, B, C, D).
• The number of permutations of 3 letters out of 4 can be calculated using the permutation formula:

$P(n, r) = \frac{n!}{(n-r)!}$

Where $n = 4$ and $r = 3$.

$P(4, 3) = \frac{4!}{(4-3)!} = \frac{4 \times 3 \times 2 \times 1!}{1!} = 4 \times 3 \times 2 = 24$

So, there are 24 ways to choose 3 letters when the order is relevant.

2. Order of the Choices is Not Relevant:

• Since the order does not matter, we use combinations.
• The number of combinations of 3 letters out of 4 can be calculated using the combination formula:

$C(n, r) = \frac{n!}{r!(n-r)!}$

Where $n = 4$ and $r = 3$.

$C(4, 3) = \frac{4!}{3!(4-3)!} = \frac{4 \times 3 \times 2 \times 1!}{3 \times 2 \times 1 \times 1!} = 4$

So, there are 4 ways to choose 3 letters when the order is not relevant.

Summary:

• When the order is relevant: 24 ways.
• When the order is not relevant: 4 ways.

Would you like more details or have any questions?

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Related Questions:

1. How would the calculation change if we had 5 letters instead of 4?
2. How can we extend this problem to choosing 2 letters instead of 3?
3. What if the letters could be chosen with replacement? How would that affect the number of ways?
4. How do the concepts of permutations and combinations apply to arranging objects in a line?
5. How would the results change if we considered choosing all 4 letters?

Tip: To determine whether to use permutations or combinations, ask yourself if the order in which you select the items makes a difference in the final outcome.