To solve this problem, let's break it down:

The student is required to answer the first and the last question. This means that these two questions are already chosen.

So, the task reduces to selecting the remaining three questions from the 7 middle questions (i.e., the questions numbered 2 through 8).

The number of ways to choose 3 questions from 7 is a combination problem. The formula for combinations is:

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

Where:

• $n = 7$ (the 7 middle questions)
• $r = 3$ (the 3 additional questions to be selected)

Now, applying this formula:

$\binom{7}{3} = \frac{7!}{3!(7 - 3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

Thus, there are 35 ways to choose the remaining 3 questions.

Final Answer:

The student can select the 5 questions in 35 ways.

Would you like to dive deeper into combinations or any other concepts related to this problem?

Here are a few related questions:

1. What is the formula for combinations, and how does it differ from permutations?
2. How would the answer change if the student were required to answer the first 3 questions and the last 2 questions?
3. What if the student could skip the first and last questions and select all 5 from the middle?
4. How do you calculate the number of ways to arrange 5 out of 9 questions in order (permutations)?
5. Can this problem be generalized to more or fewer questions, and how would the solution change?

Tip: When you're working with combinations, always remember that the order of selection does not matter, which is why you divide by the factorial of the number of items selected.