How many 3 digit even numbers can be formed from the digits 1, 2 , 3, 4, 5 and 6 assuming that repetition of the digits is allowed .

Options

A. 120
B. 96
C. 24
D. 420

Explanation

To solve the problem of how many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, and 6, with repetition of digits allowed, we can follow these steps:

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

### Step 2: Determine the condition for an even number
For a number to be even, its last digit (the unit's place) must be an even digit. From our available digits, the even digits are 2, 4, and 6.

### Step 3: Count the choices for the unit's place
Since we have 3 even digits (2, 4, 6), there are 3 choices for the unit's place.

### Step 4: Count the choices for the tens place
The tens place can be filled with any of the 6 available digits (1, 2, 3, 4, 5, 6), since repetition is allowed. Therefore, there are 6 choices for the tens place.

### Step 5: Count the choices for the hundreds place
Similarly, the hundreds place can also be filled with any of the 6 available digits (1, 2, 3, 4, 5, 6), as repetition is allowed. Thus, there are 6 choices for the hundreds place.

### Step 6: Calculate the total number of combinations
Using the fundamental principle of counting, the total number of 3-digit even numbers can be calculated by multiplying the number of choices for each place:

\[ \text{Total combinations} = (\text{choices for hundreds place}) \times (\text{choices for tens place}) \times (\text{choices for unit's place}) \]
\[ \text{Total combinations} = 6 \times 6 \times 3 \]

### Step 7: Perform the calculation
\[ \text{Total combinations} = 6 \times 6 = 36 \]
\[ \text{Total combinations} = 36 \times 3 = 108 \]

Thus, the total number of 3-digit even numbers that can be formed is **108**.

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