Suppose we want to choose 

3

 letters, without replacement, from the 

4

 letters A, B, C, and D.

(If necessary, consult a list of formulas.)

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?

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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.

Suppose we want to choose 3 letters, without replacement, from the 4 letters A, B, C, and D.

Learn how to choose 3 letters from 4 (A, B, C, D) without replacement using the combination formula C(n, r) = n! / (r!(n - r)!). This problem covers essential concepts in combinatorics, such as combinations, and provides a step-by-step explanation. Suitable for students in Grades 9-12.

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