Let's solve both problems step by step.

Problem 1:

Question: Determine the number of five-digit numbers that can be formed using the digits $\{1, 3, 5, 7, 8, 9\}$, given that:

• The first digit must be 7.
• One of the other digits must be 1.
• No digit can appear more than once.

Solution:

1. Fix the first digit as 7: Since the first digit is fixed as $7$, we now need to fill the remaining 4 positions with the digits $\{1, 3, 5, 8, 9\}$, and we must include 1 in one of those positions.

2. Choose a position for 1: We can place the digit 1 in any one of the remaining 4 positions. The number of ways to do this is: $4 \text{ ways}$

3. Fill the remaining 3 positions: After placing 1, we are left with 4 digits $\{3, 5, 8, 9\}$, and we need to fill 3 positions with no repetition. The number of ways to fill these 3 positions is given by the number of permutations of 4 digits taken 3 at a time, which is: $P(4, 3) = 4 \times 3 \times 2 = 24 \text{ ways}$

4. Total number of valid numbers: Multiply the number of ways to place 1 by the number of ways to fill the remaining digits: $4 \times 24 = 96$

So, the total number of five-digit numbers that can be formed is $\boxed{96}$.

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Problem 2:

Question: Find the number of five-digit numbers that can be formed using the digits $\{1, 2, 3, 6, 9\}$, given that:

• The first digit must be 1.
• One of the other digits must be 2.
• No digit can appear more than once.

Solution:

1. Fix the first digit as 1: The first digit is already fixed as 1, so we need to fill the remaining 4 positions with the digits $\{2, 3, 6, 9\}$, and we must include 2 in one of those positions.

2. Choose a position for 2: We can place the digit 2 in any one of the remaining 4 positions. The number of ways to do this is: $4 \text{ ways}$

3. Fill the remaining 3 positions: After placing 2, we are left with 3 digits $\{3, 6, 9\}$, and we need to fill 3 positions with no repetition. The number of ways to arrange these 3 digits is given by: $P(3, 3) = 3! = 3 \times 2 \times 1 = 6 \text{ ways}$

4. Total number of valid numbers: Multiply the number of ways to place 2 by the number of ways to arrange the remaining digits: $4 \times 6 = 24$

So, the total number of five-digit numbers that can be formed is $\boxed{24}$.

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Do you want further details or have any questions? Here are 5 related questions to expand on these problems:

1. How do permutations differ from combinations in these types of problems?
2. What if repetition of digits were allowed in either problem—how would that change the solution?
3. Can you generalize the formula to find the number of n-digit numbers with similar restrictions?
4. How would the solution change if the first digit were not fixed in either problem?
5. What other types of constraints can be introduced to change the structure of such problems?

Tip: Always clearly identify fixed positions or digits in combinatorial problems to break the problem into manageable parts.