To solve this problem, let's break it down step by step.

Step 1: Treat the three Maths books as one unit

Since the three Maths books must remain adjacent and in the same order, we can treat these three books as one single unit. This reduces the total number of units to arrange from 7 to 5 (the 4 other books + 1 unit of Maths books).

Step 2: Calculate the number of arrangements for the units

The number of ways to arrange 5 units is given by the factorial of 5, which is:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Step 3: Consider the internal arrangement of the Maths books

Since the Maths books must remain in the same order, there's only 1 way to arrange them within their unit.

Step 4: Multiply the results

Since the Maths books can only be arranged in one specific way within their unit, the total number of ways to arrange all the books is simply the number of ways to arrange the 5 units:

$120 \times 1 = 120$

So, the correct answer is:

c. 120

Would you like more details on any step, or do you have any questions?

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Here are five related questions:

1. How many ways can Amal arrange the books if there are no restrictions?
2. What if the Maths books don't have to be in the same order but still need to be adjacent?
3. How many arrangements are possible if only two of the Maths books must be adjacent?
4. What is the probability that the Maths books are adjacent if the books are arranged randomly?
5. How does the number of arrangements change if there were more or fewer books?

Tip: Factorials are useful in counting problems where order matters.