How many 9-digit numbers can be created using the digits from the number 214432213, such that the even places only accommodate the even digits in the resultant number? [units place is an odd place, tens place is an even place and so on]

To solve this problem, let's analyze the constraints step-by-step.

### Given Data and Observations

1. **Available Digits**: The number provided is **214432213**. 
   - We have the digits: \(2, 1, 4, 4, 3, 2, 2, 1, 3\).
   - **Frequency of each digit**:
     - **2** appears **3 times**.
     - **1** appears **2 times**.
     - **4** appears **2 times**.
     - **3** appears **2 times**.

2. **Objective**: Construct a **9-digit number** such that:
   - Only **even digits** occupy the **even places** (2nd, 4th, 6th, and 8th).
   - The **odd places** (1st, 3rd, 5th, 7th, and 9th) can be occupied by either even or odd digits.

3. **Identifying Even and Odd Digits**:
   - **Even digits** are **2** and **4**.
   - **Odd digits** are **1** and **3**.

4. **Even Places Requirement**:
   - Since there are four even places (2nd, 4th, 6th, and 8th), we need to fill these with the even digits.
   - We have a total of **5 even digits** (three 2's and two 4's), which is sufficient to fill the four even places.

### Step 1: Choose Digits for Even Places

Since there are **5 available even digits** and **4 even positions**, we need to:
- Select **4 out of these 5 even digits**.
  
The combinations for this are:
\[
\binom{5}{4} = 5
\]
After choosing 4 specific digits, we can arrange them in the 4 even places in **4!** ways.

Thus, for the even places:
\[
5 \times 4! = 5 \times 24 = 120 \text{ ways}
\]

### Step 2: Arrange Digits in Odd Places

After filling the even places with 4 chosen even digits, we are left with **5 remaining digits**. These will include:
- Any leftover even digit (from the original selection of 5 even digits).
- All odd digits (total of 4 odd digits initially).

Since we have **5 remaining digits** for the **5 odd places** (1st, 3rd, 5th, 7th, 9th), we can arrange them in **5!** ways.

Thus, for the odd places:
\[
5! = 120 \text{ ways}
\]

### Step 3: Total Number of Valid 9-Digit Numbers

Now, we can find the total number of valid 9-digit numbers by multiplying the number of ways to arrange digits in the even places and the odd places:
\[
120 \times 120 = 14400
\]

### Conclusion

The total number of 9-digit numbers that can be created under the given constraints is:
\[
\boxed{14400}
\]

Would you like further clarification on any part of the solution?

---

Here are 5 additional questions to explore:
1. What if the number had a different distribution of even and odd digits?
2. How would the solution change if there were only 3 even places?
3. What if we required specific digits in certain positions?
4. How does this type of combinatorial problem generalize to numbers of different lengths?
5. How would the solution be affected if duplicates of each digit were not allowed?

**Tip**: In combinatorics, breaking down complex constraints into independent, smaller parts can simplify counting.

Explore how to determine the number of 9-digit numbers using the digits in 214432213 with specific constraints on even and odd placements. This solution covers combinatorial concepts, including combinations, permutations, and the Fundamental Counting Principle. Ideal for high school students in Grades 10-12.

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