To solve the problem of choosing 3 letters from 4 letters (A, B, C, D) without replacement, we can use the combination formula. A combination is used when the order of selection doesn't matter.

The formula for combinations is:

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

Where:

• $n$ is the total number of items to choose from (in this case, 4 letters),
• $r$ is the number of items to choose (in this case, 3 letters),
• $n!$ denotes the factorial of $n$, which is the product of all positive integers up to $n$.

Now let's apply this formula:

$C(4, 3) = \frac{4!}{3!(4 - 3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = \frac{24}{6} = 4$

Thus, there are 4 different ways to choose 3 letters from 4 without replacement.

The 4 possible combinations are:

• A, B, C
• A, B, D
• A, C, D
• B, C, D

Would you like further details on this solution, or do you have any other questions?

---

Here are 5 related questions to explore further:

1. How would the solution change if the order of selection mattered (permutations)?
2. What if you had 5 letters to choose from instead of 4?
3. How would you calculate the number of ways to choose 2 letters from the 4 given?
4. Can you explain how factorials work in combinations and permutations?
5. What happens if replacement is allowed when choosing the letters?

Tip: Remember that combinations are for situations where the order does not matter, while permutations are used when the order does matter.