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Let's solve the problems step-by-step.

### Problem Breakdown
The question states:
- There are **25 students** total, with:
  - **12 engineering majors**
  - **13 math majors**
- We are selecting **8 students**.

#### Question (a)
We need to determine the number of ways to select 8 students from a group of 25, without regard to their majors.

This is a combination problem, as the order of selection does not matter. We use the combination formula:
\[
\binom{n}{r} = \frac{n!}{r!(n - r)!}
\]
where \( n = 25 \) and \( r = 8 \).

So, for question (a):
\[
\binom{25}{8} = \frac{25!}{8!(25 - 8)!} = \frac{25!}{8! \times 17!}
\]

Calculating this will give the answer for part (a).

#### Question (b)
Here, we are asked to select **5 engineering majors** and **3 math majors**.

1. First, we determine the number of ways to select 5 engineering majors from 12:
   \[
   \binom{12}{5} = \frac{12!}{5! \times (12 - 5)!} = \frac{12!}{5! \times 7!}
   \]

2. Then, we determine the number of ways to select 3 math majors from 13:
   \[
   \binom{13}{3} = \frac{13!}{3! \times (13 - 3)!} = \frac{13!}{3! \times 10!}
   \]

3. Since we need both selections (5 engineering majors and 3 math majors), we multiply these results together:
   \[
   \binom{12}{5} \times \binom{13}{3}
   \]

Let's calculate these values.

### Solutions
#### Part (a):
\[
\binom{25}{8} = 1081575
\]

#### Part (b):
1. \(\binom{12}{5} = 792\)
2. \(\binom{13}{3} = 286\)
3. Multiplying these:
   \[
   792 \times 286 = 226512
   \]

### Final Answers:
- **(a)** \( \binom{25}{8} = 1081575 \)
- **(b)** \( \binom{12}{5} \times \binom{13}{3} = 226512 \)

Would you like further details on any part of the solution? 

Here are some related questions to deepen understanding:

1. How would the answers change if there were 15 engineering and 10 math majors?
2. What if we were selecting a different number, such as 10 students instead of 8?
3. How would we approach this if the selection were ordered rather than unordered?
4. Can you calculate the probability of selecting exactly 5 engineering majors and 3 math majors from the group of 25?
5. How would we interpret these results in the context of probability or real-world selection?

**Tip:** Remember, combinations are used when the order of selection does not matter, while permutations are used when the order does matter.

A group of 25 students includes 12 engineering majors and 13 math majors. Suppose that 8 students are randomly selected.

 a. In how many ways is it possible to select 8 students?

 b. In how many ways is it possible to select 5 of the engineering majors and 3 of the math majors?

Explore how to calculate combinations in a scenario involving selecting students from a group of engineering and math majors. This problem covers combinatorics, including the combination formula and how to apply it to real-world selection problems. Suitable for high school students in Grades 10-12.

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