Let's break down and solve each part of the problem.

Part (a):

We have three groups of students, each from a different year:

• Fifth-year students: Sheila, Ann, Harvey (3 students)
• Fourth-year students: Sarah, Jeff, John (3 students)
• Third-year students: Alan, Chris, Mark (3 students)

The question asks how many ways these students can be arranged in a line if the students from each year must stay together.

1. Treat each year group as a block: Since each group must stay together, we treat each group as one "block". So, there are 3 blocks to arrange:

   • Fifth-year group
   • Fourth-year group
   • Third-year group

   The number of ways to arrange these 3 blocks is $3!$ (since there are 3 blocks to permute).

2. Arrange students within each block:

   • For the fifth-year group (Sheila, Ann, Harvey), the number of ways to arrange them within the block is $3!$.
   • For the fourth-year group (Sarah, Jeff, John), the number of ways to arrange them is also $3!$.
   • For the third-year group (Alan, Chris, Mark), the number of ways to arrange them is $3!$.

The total number of ways to arrange the students, keeping each year's group together, is: $3! \times 3! \times 3! \times 3! = 6 \times 6 \times 6 \times 6 = 1296$

Part (b):

We have six boys and two girls, and we need to find the number of ways to arrange them in a line such that the two girls do not sit together.

1. Total arrangements without restrictions: The total number of ways to arrange all 8 students (6 boys + 2 girls) is $8!$.

2. Arrangements where the two girls sit together:

   • Treat the two girls as one "block" so that they are always together. This gives us 7 "blocks" to arrange (6 boys + 1 block of 2 girls).
   • The number of ways to arrange these 7 blocks is $7!$.
   • The number of ways to arrange the two girls within their block is $2!$.

   So, the total number of ways where the two girls are together is $7! \times 2!$.

3. Subtract the cases where the girls are together from the total arrangements: $\text{Desired arrangements} = 8! - (7! \times 2!)$

Let’s compute the exact values: $8! = 40320, \quad 7! = 5040, \quad 2! = 2$ $\text{Arrangements where the girls are together} = 5040 \times 2 = 10080$ $\text{Arrangements where the girls are not together} = 40320 - 10080 = 30240$

Final Answers:

• (a) The number of ways the students can be arranged with students from each year kept together is 1296.
• (b) The number of ways to arrange six boys and two girls such that the two girls do not sit together is 30240.

Would you like more details or explanations on any step?

Here are 5 related questions:

1. How would the answer change in part (a) if there were 4 groups of students instead of 3?
2. What if we only wanted the girls to sit in alternating positions with the boys in part (b)?
3. How would part (a) change if students from different years had to alternate positions?
4. In part (b), what if there were 3 girls instead of 2?
5. Can you generalize the solution in part (b) for $n$ boys and $m$ girls?

Tip: Always try breaking down complex arrangement problems by considering restrictions first (e.g., blocking certain elements together) and then applying basic permutation rules!