How many ways are there to arrange the first ten positive integers such that the multiples of 2 appear in increasing order, and the multiples of 3 appear in decreasing order?

To solve the problem, we need to arrange the first ten positive integers \( \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \) under the following conditions:

1. The multiples of 2 (i.e., \( \{2, 4, 6, 8, 10\} \)) must appear in increasing order.
2. The multiples of 3 (i.e., \( \{3, 6, 9\} \)) must appear in decreasing order.

### Step 1: Identify the Positions
- There are 5 multiples of 2: \(2, 4, 6, 8, 10\).
- There are 3 multiples of 3: \(3, 6, 9\).
- Note that 6 is both a multiple of 2 and 3, so its position is fixed by both constraints.

### Step 2: Fix the Positions for Multiples of 2
Since the multiples of 2 must appear in increasing order, the positions of \( \{2, 4, 6, 8, 10\} \) in any valid permutation are completely determined once the positions of these numbers are chosen. If we select 5 distinct positions from the set of 10, there is only 1 way to place \( \{2, 4, 6, 8, 10\} \) in these positions in the correct order.

### Step 3: Fix the Positions for Multiples of 3
Similarly, for the multiples of 3, \( \{3, 6, 9\} \) must appear in decreasing order. Given the position of 6 (which is already fixed from Step 2), we have to choose positions for 3 and 9. As with the multiples of 2, once we choose 3 positions for \( \{3, 6, 9\} \), there is only 1 way to arrange them in the correct decreasing order.

### Step 4: Determine the Number of Ways to Arrange
- Choose 5 positions out of 10 for the multiples of 2. The number of ways to do this is given by the binomial coefficient \( \binom{10}{5} \).
- The remaining 5 positions will automatically be for the non-multiples of 2, which can be arranged in any order, giving \(5!\) arrangements.

So, the total number of valid arrangements is:
\[
\text{Total number of ways} = \binom{10}{5} \times 5!
\]

### Calculation
\[
\binom{10}{5} = \frac{10!}{5! \times 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252
\]
\[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\]

Thus, the total number of valid permutations is:
\[
252 \times 120 = 30240
\]

### Final Answer
There are **30,240** ways to arrange the first ten positive integers such that the multiples of 2 appear in increasing order and the multiples of 3 appear in decreasing order.

Would you like any further details or have any questions?

### Follow-Up Questions:
1. What if the multiples of 2 must be in decreasing order instead?
2. How would the problem change if there were more constraints on other numbers?
3. What if the multiples of 5 had to be in increasing order?
4. Can you calculate how many ways there would be if the numbers were arranged in a circle instead?
5. How would the problem change if we added more numbers beyond 10?

**Tip:** Understanding permutations and combinations is crucial for solving problems involving arrangements under specific constraints.

Explore how to arrange the first ten positive integers such that multiples of 2 appear in increasing order and multiples of 3 appear in decreasing order. Learn the detailed solution involving permutations, combinations, and binomial coefficients. Suitable for high school students.

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