To solve this problem, we need to consider two scenarios:

1. Order of the Choices is Relevant:

• Since the order matters, we are dealing with permutations.
• We are choosing 2 letters from a set of 5 letters (A, B, C, D, E).
• The number of permutations of 2 letters out of 5 can be calculated using the permutation formula:

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

Where $n = 5$ (total letters) and $r = 2$ (letters to choose).

$P(5, 2) = \frac{5!}{(5-2)!} = \frac{5 \times 4 \times 3!}{3!} = 5 \times 4 = 20$

So, there are 20 ways to choose 2 letters when the order is relevant.

2. Order of the Choices is Not Relevant:

• Since the order does not matter, we are dealing with combinations.
• The number of combinations of 2 letters out of 5 can be calculated using the combination formula:

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

Where $n = 5$ and $r = 2$.

$C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{5 \times 4}{2 \times 1} = 10$

So, there are 10 ways to choose 2 letters when the order is not relevant.

Summary:

• When the order is relevant: 20 ways.
• When the order is not relevant: 10 ways.

Would you like more details or have any questions?

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Related Questions:

1. What is the difference between permutations and combinations?
2. How would the calculation change if we had 6 letters instead of 5?
3. How can we extend this problem to choosing 3 letters instead of 2?
4. What if we wanted to choose the letters with replacement? How would that affect the number of ways?
5. How does this problem relate to real-world scenarios, like password creation or lottery numbers?

Tip: Always remember, permutations are used when the order matters, while combinations are used when the order does not matter.