This problem can be solved using the concept of permutations because the order of the letters matters and no letter is repeated.

Explanation:

We are choosing and arranging 4 letters from a set of 7 letters (A, B, C, D, E, F, G), without repetition. The number of such arrangements is given by:

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

Where:

• $n$ is the total number of items (7 letters),
• $r$ is the number of items to arrange (4 letters).

Substitute $n = 7$ and $r = 4$:

$P(7, 4) = \frac{7!}{(7 - 4)!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!}$

Cancel out $3!$:

$P(7, 4) = 7 \times 6 \times 5 \times 4 = 840$

Final Answer:

There are 840 different codes possible.

Would you like to see further details or examples of similar problems?

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Here are 5 related questions to explore:

1. What if the letters can be repeated? How many codes would be possible then?
2. How many codes would be possible if only 3 letters were used instead of 4?
3. What is the probability of guessing the correct code on the first attempt?
4. How does the number of codes change if there are 8 letters (A-H) available instead of 7?
5. What is the total number of combinations (not arrangements) of 4 letters from A-G?

Tip: For problems involving permutations, always consider whether repetition is allowed or not—it changes the calculation significantly!