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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?

Selesaikan soal-soal berikut: 1. Dari kota A ke kota B dapat ditempuh dengan 2 cara, dari kota B ke kota C dapat ditempuh dengan 4 cara. Tentukan banyaknya cara yang dapat ditempuh... (other questions follow).

Solve a variety of combinatorics problems involving traveling routes, number formations, and word arrangements. Learn how to calculate using permutations with repetition and combinations. Explore problems that involve counting methods and quadratic equations.

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