Consider a collection of cards numbered from 1 to 9.
1 2 3
4 5 6
7 8 9

All nine cards are arranged in a row so that if the numbers are read from
left to right, the odd numbers are in increasing order. The even-numbered
cards can be in any order. An example is shown below.
1 8 4 3 5 7 2 6 9                                                                                                                                      How many ways can eight of these cards be selected and then arranged
in a row so that the odd numbers are in increasing order?

To solve the problem, we first need to analyze the requirements:

### Problem Breakdown:
1. We have nine cards numbered from 1 to 9.
2. The cards are arranged in a row such that the odd numbers (1, 3, 5, 7, 9) must be in increasing order.
3. The even-numbered cards (2, 4, 6, 8) can be in any order.
4. We need to find the number of ways to select and arrange **eight** cards so that the odd numbers are in increasing order.

### Step 1: Identify the Possible Scenarios
Since we are selecting eight cards out of the nine, we can consider which card is excluded. The excluded card could be:
- One of the odd numbers: 1, 3, 5, 7, 9 (5 possibilities)
- One of the even numbers: 2, 4, 6, 8 (4 possibilities)

### Case 1: Excluding an Odd Number
- If one odd number is excluded, say \(x\), then the remaining odd numbers must be in increasing order. The position of these remaining odd numbers is fixed.
- The number of remaining even cards is 4, which can be arranged in the 4 remaining spots in any order.

For example, if 1 is excluded, then the sequence must look like \(3, 5, 7, 9\) in that order, with the four even cards filling the remaining spaces.

For each choice of an odd number to exclude (5 choices), the 4 even numbers can be arranged in the 4 remaining spots in \(4!\) ways. So, the total number of arrangements for this case is:
\[
5 \times 4! = 5 \times 24 = 120 \text{ ways}
\]

### Case 2: Excluding an Even Number
- If one even number is excluded, then the five odd numbers must still be in their fixed increasing order.
- The remaining three even cards can be placed in any of the 3 remaining spots.

For each choice of an even number to exclude (4 choices), the 3 remaining even numbers can be arranged in the 3 remaining spots in \(3!\) ways. So, the total number of arrangements for this case is:
\[
4 \times 3! \times \binom{8}{3} = 4 \times 6 \times 56 = 1344 \text{ ways}
\]

### Total Number of Ways
Finally, the total number of ways to arrange the eight cards, considering both cases, is:
\[
120 + 1344 = 1464 \text{ ways}
\]

Thus, the total number of ways to select and arrange eight of these cards so that the odd numbers are in increasing order is **1464**.

Would you like a detailed breakdown of each step? Here are some related questions you might find interesting:

1. What if the cards were numbered from 1 to 10, with similar conditions?
2. How would the problem change if we had to exclude two cards instead of one?
3. Can you describe how the arrangements would differ if the odd numbers had to be in *decreasing* order?
4. How would you calculate the number of arrangements if there were more even cards than odd ones?
5. Could this problem be generalized to cards numbered from 1 to any arbitrary \( n \)?

**Tip:** When working with combinatorial problems, breaking down the problem into manageable cases is often a helpful strategy.

Explore how to arrange eight cards numbered 1 to 9 with increasing odd numbers and any order for even numbers. Learn about combinatorial arrangements and factorials. Suitable for advanced high school students.

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