To determine how many 3-letter code words can be formed using the first 8 letters of the alphabet (A, B, C, D, E, F, G, H) with no repeats allowed, we need to consider the number of available choices for each letter in the code word.

• For the first letter, there are 8 possible choices.
• Once the first letter is chosen, only 7 letters remain available for the second position.
• After choosing the second letter, only 6 letters remain for the third position.

Therefore, the total number of possible 3-letter code words is given by multiplying the number of choices for each position:

$8 \times 7 \times 6 = 336$

Thus, 336 different 3-letter code words are possible.

Would you like more details, or have any questions?

Here are 5 related questions to consider:

1. How many 3-letter code words are possible if repeats are allowed?
2. How many 4-letter code words can be formed using the first 8 letters with no repeats?
3. How many 3-letter code words can be formed using only vowels from the first 8 letters (A, E)?
4. What is the probability of randomly selecting a code word that starts with 'A'?
5. If we restrict the code words to only even-positioned letters in the alphabet (B, D, F, H), how many are possible?

Tip: When dealing with problems involving choices, multiplying the available options step-by-step helps count all possible outcomes efficiently!