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
(a) How many ways can this be done?

Let's analyze the problem step by step.

### Given:
- A collection of 9 cards numbered from 1 to 9.
- The odd-numbered cards are \(1, 3, 5, 7, 9\), and they must be in increasing order.
- The even-numbered cards are \(2, 4, 6, 8\), and they can be in any order.

### Objective:
To find out how many ways the cards can be arranged such that the odd numbers are in increasing order.

### Solution:

1. **Odd-numbered cards:**
   - The odd numbers must appear in the sequence \(1, 3, 5, 7, 9\). The positions for these numbers must be selected in such a way that they maintain this order.
   - There are 5 odd-numbered cards, and we need to choose 5 positions out of the 9 available slots to place these cards. The number of ways to choose 5 positions from 9 is given by the binomial coefficient \( \binom{9}{5} \).

2. **Even-numbered cards:**
   - The remaining 4 positions will automatically be assigned to the even-numbered cards.
   - The even-numbered cards \(2, 4, 6, 8\) can be arranged in any order in these 4 positions. The number of ways to arrange 4 items is \(4!\).

### Total Number of Arrangements:
- The total number of valid arrangements is the product of the number of ways to select positions for the odd-numbered cards and the number of ways to arrange the even-numbered cards in the remaining positions.

\[
\text{Total Arrangements} = \binom{9}{5} \times 4!
\]

Now, let's calculate each part:

\[
\binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5! \times 4 \times 3 \times 2 \times 1} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126
\]

\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]

\[
\text{Total Arrangements} = 126 \times 24 = 3024
\]

### Final Answer:
There are **3024** different ways to arrange the cards such that the odd numbers are in increasing order.

Would you like any further details or have any questions?

### Related Questions:
1. What if the cards are numbered from 1 to 10? How does that change the total arrangements?
2. How many ways can the even-numbered cards be arranged if they must also be in increasing order?
3. What if only 3 odd cards need to be in increasing order out of a set of 9 cards?
4. How would the problem change if duplicates of numbers were allowed?
5. How many ways are there to arrange the cards if both odd and even numbers must be in increasing order?

### Tip:
Understanding the use of binomial coefficients and factorials is essential for counting arrangements in combinatorial problems.

Explore how to count the number of ways to arrange cards numbered from 1 to 9 such that odd numbers are in increasing order. This combinatorial problem involves selecting positions for odd numbers and arranging even numbers. Learn with step-by-step calculations and examples.

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