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.