consider a collection of cards numbered from 1 to 9. 1 23 4 5 6 789 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 26 9 (a) how many ways can this be done choosing only 8 cards?

We are given a collection of 9 cards numbered from 1 to 9, and the task is to arrange 8 of these cards so that:

1. The odd-numbered cards are arranged in increasing order.
2. The even-numbered cards can be arranged in any order.

### Step-by-Step Approach:

#### Step 1: Identify Odd and Even Numbers
From the cards numbered 1 to 9:
- **Odd-numbered cards**: \( \{1, 3, 5, 7, 9\} \)
- **Even-numbered cards**: \( \{2, 4, 6, 8\} \)

Since we are selecting only 8 cards out of 9, either one odd-numbered card or one even-numbered card will be left out.

#### Step 2: Consider Different Cases
Let's analyze the problem by considering two cases:

- **Case 1**: One odd-numbered card is excluded.
- **Case 2**: One even-numbered card is excluded.

---

### Case 1: Excluding One Odd-Numbered Card
If one odd-numbered card is excluded, we will use 4 odd cards in total, which must be arranged in increasing order (so their arrangement is fixed). The remaining 4 cards will be chosen from the even-numbered cards.

1. **Choose 4 odd-numbered cards**: We can exclude any one of the 5 odd-numbered cards. There are 5 choices for which odd-numbered card is excluded.
2. **Arrange the even-numbered cards**: The 4 even-numbered cards can be arranged in any order. The number of ways to arrange 4 even-numbered cards is \( 4! \).

So, the total number of arrangements in this case is:
\[
5 \times 4! = 5 \times 24 = 120
\]

---

### Case 2: Excluding One Even-Numbered Card
If one even-numbered card is excluded, we will use all 5 odd-numbered cards, which are arranged in increasing order. Then we need to choose and arrange 3 out of the 4 even-numbered cards.

1. **Choose 3 even-numbered cards**: The number of ways to choose 3 out of 4 even-numbered cards is \( \binom{4}{3} = 4 \).
2. **Arrange the chosen even-numbered cards**: The 3 even-numbered cards can be arranged in \( 3! \) ways.

So, the total number of arrangements in this case is:
\[
4 \times 3! = 4 \times 6 = 24
\]

---

### Final Answer: Add Both Cases
The total number of ways to arrange 8 cards is the sum of the two cases:
\[
120 + 24 = 144
\]

Thus, the number of ways to arrange 8 cards such that the odd-numbered cards are in increasing order and the even-numbered cards can be in any order is **144**.

---

Would you like further details or clarifications?

### Related Questions:
1. How many ways can the entire set of 9 cards be arranged under the same conditions?
2. What if there are 10 cards numbered from 1 to 10, and you arrange 8 of them with the same conditions?
3. How would the problem change if both odd and even numbers had to be in increasing order?
4. Can you generalize this method to any set of odd and even numbers?
5. How many ways can 7 cards be arranged with 4 odd and 3 even numbers?

### Tip:
When the arrangement of certain elements is fixed (like the odd numbers in increasing order), the key focus is on how to select and arrange the other elements (like the even numbers).

Explore how to arrange 8 cards numbered from 1 to 9, where odd-numbered cards must be in increasing order and even-numbered cards can be in any order. This combinatorics problem includes factorials, permutations, and combinations. Suitable for students in Grades 9-12.

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