Let's work through each of the problems step by step:

1. Traveling between cities A, B, and C:

You are given the following:

• From city A to city B: 2 possible ways.
• From city B to city C: 4 possible ways.

a. From city A to city C via B: You multiply the two options (since they are independent): $2 \times 4 = 8 \text{ ways}$

b. From city A to city C and back to city A (without restrictions): You need to account for both the outgoing and return trips. For the outgoing trip: $2 \times 4 = 8 \text{ ways}$ For the return trip (since it's independent): $2 \times 4 = 8 \text{ ways}$ So the total number of round trips is: $8 \times 8 = 64 \text{ ways}$

c. From city A to city C and back to city A without repeating routes: For the outward journey, there are 8 ways (as before). However, for the return trip, the restriction is that the same route cannot be used, so you have 7 remaining ways for the return journey. Thus: $8 \times 7 = 56 \text{ ways}$

2. Number of numbers between 450 and 700:

The digits available are 2, 3, 4, 5, 6, and 7. The number must be three digits, starting with either 4, 5, or 6 to be between 450 and 700.

For hundreds place:

• You can use 4, 5, or 6 (3 options).

For tens and units places:

• You can choose from 2, 3, 5, 6, 7 (5 options for each).

So, the total number of such numbers is: $3 \times 5 \times 5 = 75 \text{ numbers}$

3. Counting letter arrangements:

a. GAGAL TOTAL: This word has 10 letters, with the following repetitions:

• G appears 2 times.
• A appears 3 times.
• L appears 2 times.

The formula for counting arrangements with repetitions is: $\frac{10!}{2!3!2!}$

Calculating this: $\frac{10!}{2!3!2!} = \frac{3628800}{2 \times 6 \times 2} = \frac{3628800}{24} = 151200$

b. MAMA MEMASAK: This word has 11 letters, with the following repetitions:

• M appears 4 times.
• A appears 4 times.
• E appears 2 times.

The number of arrangements is: $\frac{11!}{4!4!2!}$

Calculating: $\frac{11!}{4!4!2!} = \frac{39916800}{24 \times 24 \times 2} = \frac{39916800}{1152} = 34650$

4. Solving for $n$ in:

a. $\frac{n}{2} \times \left(\frac{n}{2} - 1\right) = 42$

Let’s expand this: $\frac{n}{2} \times \frac{n-2}{2} = 42$ $\frac{n(n-2)}{4} = 42$ $n(n-2) = 168$ Solve this quadratic: $n^2 - 2n - 168 = 0$ Using the quadratic formula: $n = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-168)}}{2(1)}$ $n = \frac{2 \pm \sqrt{4 + 672}}{2} = \frac{2 \pm \sqrt{676}}{2} = \frac{2 \pm 26}{2}$ So $n = 14$ or $n = -12$. Since $n$ must be positive, we take $n = 14$.

b. $6! = (n+3)! = 4!(n+4)!$: This equation involves factorials and would require solving based on pattern matching for small values of $n$.

Would you like to proceed with any specific question, or should I continue with more detailed solutions?