To solve the problem of finding the total number of arrangements of the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 such that the odd digits occupy even places, we can follow these steps:
### Step 1: Identify the digits and their counts
We have the digits:
- 1 (two times)
- 2 (four times)
- 3 (one time)
- 4 (two times)
Thus, the digits are 1, 1, 2, 2, 2, 2, 3, 4, 4.
### Step 2: Determine the positions available
Since we have 9 digits, the positions are numbered from 1 to 9. The even positions are 2, 4, 6, and 8, which gives us a total of 4 even positions.
### Step 3: Identify the odd digits
The odd digits present are:
- 1 (two times)
- 3 (one time)
So we have three odd digits in total: 1, 1, and 3.
### Step 4: Place odd digits in even positions
We need to place the three odd digits (1, 1, 3) in the 4 available even positions (2, 4, 6, 8). We can choose 3 positions out of the 4 for the odd digits.
The number of ways to choose 3 positions from 4 is given by the combination formula:
\[
\binom{4}{3} = 4
\]
### Step 5: Arrange the odd digits in the chosen positions
Now, we need to arrange the selected odd digits (1, 1, 3) in the chosen positions. The arrangement can be calculated as:
\[
\frac{3!}{2!} = 3
\]
This accounts for the repetition of the digit '1'.
### Step 6: Place the even digits in the remaining positions
After placing the odd digits, we have 6 positions left (1, 3, 5, 7, 9) for the even digits (2, 2, 2, 2, 4, 4).
The number of ways to arrange the even digits (2, 2, 2, 2, 4, 4) in these 6 positions is given by:
\[
\frac{6!}{4! \cdot 2!}
\]
### Step 7: Calculate the total arrangements
Now we can combine all the calculations:
1. Choose positions for odd digits: \( \binom{4}{3} = 4 \)
2. Arrange odd digits: \( \frac{3!}{2!} = 3 \)
3. Arrange even digits: \( \frac{6!}{4! \cdot 2!} = \frac{720}{24 \cdot 2} = 15 \)
Thus, the total number of arrangements is:
\[
\text{Total} = 4 \times 3 \times 15 = 180
\]
### Final Answer
The total number of arrangements where odd digits occupy even places is **180**.
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