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Let's break down the problem to solve each part step by step.

We are given that \( n = 7 \), meaning we are using the first 7 letters of the alphabet: \(\{ a, b, c, d, e, f, g \}\), to build 5-letter "words" (strings).

### Part (a) - How many of these words are there total?
For each of the 5 positions in the word, we can choose any of the 7 letters, and repetition is allowed. Therefore, the total number of possible 5-letter words is:

\[
7^5 = 16807
\]

### Part (b) - How many of these words contain no repeated letters?
If no letters are repeated, we are choosing 5 distinct letters from the 7 available letters. This is a permutation problem, as the order matters.

The number of such words is:

\[
7 \times 6 \times 5 \times 4 \times 3 = 2520
\]

### Part (c) - How many of these words start with the sub-word "ade"?
If the word starts with "ade", then the first three letters are fixed as "a", "d", and "e". For the remaining two letters, we can choose from any of the 7 letters (with repetition allowed). Therefore, the number of such words is:

\[
7 \times 7 = 49
\]

### Part (d) - How many of these words either start with "ade" or end with "be" or both?
1. Words that start with "ade" – We have already calculated this in part (c): \( 49 \).
2. Words that end with "be" – If the word ends with "be", then the last two letters are fixed as "b" and "e", and we can choose the first three letters freely (with repetition allowed), so the number of such words is:

\[
7 \times 7 \times 7 = 343
\]

3. Words that both start with "ade" and end with "be" – If both conditions are true, the word is of the form "ade**be**", so the middle part is fixed. There is exactly 1 such word.

To get the total, we use the principle of inclusion and exclusion:
\[
49 \, (\text{starts with "ade"}) + 343 \, (\text{ends with "be"}) - 1 \, (\text{both}) = 391
\]

### Part (e) - How many of the words containing no repeats also do not contain the sub-word "bed"?
1. First, calculate how many words have no repeated letters, which we did in part (b): \( 2520 \).
2. Now, count how many of these words contain "bed" as a sub-word. If "bed" appears as a sub-word, then the positions of "b", "e", and "d" are fixed, and the remaining two positions must be filled with the remaining 4 letters from \(\{a, c, f, g\}\), without repetition. The number of such words is:

\[
4 \times 3 = 12
\]

Thus, the number of words containing no repeats and also not containing "bed" is:

\[
2520 - 12 = 2508
\]

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### Summary of answers:
a) Total number of words: \( 16807 \)  
b) Words with no repeated letters: \( 2520 \)  
c) Words starting with "ade": \( 49 \)  
d) Words starting with "ade" or ending with "be" or both: \( 391 \)  
e) Words with no repeats and without "bed": \( 2508 \)

Do you need further details or have any questions?

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Here are 5 follow-up questions to explore further:
1. How would the results change if \(n = 8\) instead of \(n = 7\)?
2. What if repetition of letters was not allowed at all in the problem?
3. Can you extend the problem to 6-letter words?
4. How would the results change if we restricted words to only even positions being vowels?
5. Can you apply similar counting techniques to a different alphabet, say binary digits?

**Tip:** For problems involving counting, always consider whether repetition is allowed and whether the order matters—these two factors often decide whether to use combinations or permutations!

We want to build 5 letter 'words' using only the first n = 7 letters of the alphabet. For example, in n = 5 we can use the first 5 letters, {a, b, c, d, e}. (Recall, words are just strings of letters, not necessarily actual English words.)

Explore the combinatorial analysis of constructing 5-letter words using the first 7 letters of the alphabet, including total possibilities, restrictions on letter repetition, and specific substrings. This study incorporates concepts such as permutations and the inclusion-exclusion principle.

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