To solve this problem, we begin by understanding the conditions given:

Step 1: Identify the odd and even cards

The odd-numbered cards from 1 to 9 are: $1, 3, 5, 7, 9$ The even-numbered cards from 1 to 9 are: $2, 4, 6, 8$

Step 2: Fix the positions of the odd cards

Since the odd numbers must be in increasing order from left to right, their order is fixed as $1, 3, 5, 7, 9$. However, we need to choose which positions in the row of 9 cards will be occupied by these odd numbers.

Step 3: Choose positions for odd cards

There are 9 positions available in the row, and we need to choose 5 positions for the odd cards. The number of ways to choose 5 positions out of 9 is given by the binomial coefficient: $\binom{9}{5} = \binom{9}{4} = 126$

Step 4: Arrange the even cards

The remaining 4 positions will be occupied by the even cards $2, 4, 6, 8$. These can be arranged in any order. The number of ways to arrange 4 distinct cards is: $4! = 24$

Step 5: Calculate the total number of arrangements

To find the total number of arrangements, multiply the number of ways to choose the positions of the odd cards by the number of ways to arrange the even cards: $126 \times 24 = 3024$

Final Answer

The total number of ways the cards can be arranged is: $\boxed{3024}$

Would you like more details or have any questions?

Related Questions:

1. How many ways can the even numbers be arranged if they had to be in increasing order as well?
2. What if the odd numbers could also be in any order? How would the answer change?
3. How would the total number of arrangements change if there were 10 cards instead of 9?
4. How many ways can the cards be arranged if both odd and even numbers need to be in decreasing order?
5. How many ways can the arrangement be done if we have an additional restriction, such as even numbers cannot be next to each other?

Tip:

When solving combinatorial problems, always start by fixing any required ordering or restrictions, as this simplifies calculations significantly.