How many five - figures numbers can be formed with the digits ,1,2,3,4,5 given that an even digit may not occupy an even place if the digits may be repeated ?

Options

A. a. 1125
B. b. 5210
C. c. 1152
D. d. 2120

Answer

To solve the problem of how many five-figure numbers can be formed with the digits 1, 2, 3, 4, and 5, given that even digits may not occupy an even place, we will follow these steps:

### Step 1: Identify the digits and their classifications
The digits we have are: 1, 2, 3, 4, 5.
- **Even digits:** 2, 4
- **Odd digits:** 1, 3, 5

### Step 2: Identify the positions in the five-figure number
In a five-digit number, the positions are as follows:
1. Position 1 (odd)
2. Position 2 (even)
3. Position 3 (odd)
4. Position 4 (even)
5. Position 5 (odd)

### Step 3: Determine the allowed digits for each position
According to the problem, even digits (2, 4) cannot occupy even positions (2 and 4). Therefore:
- **Positions 1, 3, and 5 (odd positions)** can have any digit (1, 2, 3, 4, 5).
- **Positions 2 and 4 (even positions)** can only have odd digits (1, 3, 5).

### Step 4: Calculate the number of choices for each position
- **For positions 1, 3, and 5 (odd positions):** Each can be filled with any of the 5 digits (1, 2, 3, 4, 5).
- Choices for position 1: 5
- Choices for position 3: 5
- Choices for position 5: 5

- **For positions 2 and 4 (even positions):** Each can only be filled with odd digits (1, 3, 5).
- Choices for position 2: 3
- Choices for position 4: 3

### Step 5: Calculate the total number of five-figure numbers
The total number of combinations can be calculated by multiplying the number of choices for each position:
\[ \text{Total} = (\text{Choices for position 1}) \times (\text{Choices for position 2}) \times (\text{Choices for position 3}) \times (\text{Choices for position 4}) \times (\text{Choices for position 5}) \]
\[ \text{Total} = 5 \times 3 \times 5 \times 3 \times 5 \]

### Step 6: Perform the multiplication
Calculating the above expression:
\[ \text{Total} = 5 \times 3 = 15 \]
\[ \text{Total} = 15 \times 5 = 75 \]
\[ \text{Total} = 75 \times 3 = 225 \]
\[ \text{Total} = 225 \times 5 = 1125 \]

### Final Answer
The total number of five-figure numbers that can be formed is **1125**.

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