How many 3-digit even numbers can be formed from the digit 1,2,3,4,5,6, 9 if the digits can be repeated ?

Answer

To solve the problem of how many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6, and 9 with repetition allowed, we can break it down step by step.

### Step 1: Identify the digits
The available digits are: 1, 2, 3, 4, 5, 6, 9.

### Step 2: Determine the last digit
Since we want to form an even number, the last digit (unit place) must be an even digit. From the available digits, the even digits are: 2, 4, and 6.

Thus, we have **3 choices** for the last digit.

### Step 3: Determine the first digit
The first digit of a 3-digit number can be any of the available digits (1, 2, 3, 4, 5, 6, 9). Since repetition is allowed, we have **7 choices** for the first digit.

### Step 4: Determine the second digit
Similarly, the second digit can also be any of the available digits (1, 2, 3, 4, 5, 6, 9). Again, since repetition is allowed, we have **7 choices** for the second digit.

### Step 5: Calculate the total combinations
Now, we can calculate the total number of 3-digit even numbers by multiplying the number of choices for each digit:

\[ \text{Total combinations} = (\text{choices for first digit}) \times (\text{choices for second digit}) \times (\text{choices for last digit}) \]

Substituting the values we found:

\[ \text{Total combinations} = 7 \times 7 \times 3 \]

Calculating this gives:

\[ 7 \times 7 = 49 \]
\[ 49 \times 3 = 147 \]

Therefore, the total number of 3-digit even numbers that can be formed is **147**.

### Summary of Steps
1. Identify available digits.
2. Determine choices for the last digit (even).
3. Determine choices for the first digit.
4. Determine choices for the second digit.
5. Multiply the number of choices to find the total combinations.