How many 6 digit number can be formed from the digits 1, 2, 3, 4, 5, 6 which are divisible by 4 and digits are not repeated

Options

A. 192
B. 122
C. 140
D. 242

Answer

To find how many 6-digit numbers can be formed from the digits 1, 2, 3, 4, 5, and 6 that are divisible by 4 and have no repeated digits, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the last two digits for divisibility by 4**:
A number is divisible by 4 if the number formed by its last two digits is divisible by 4. We need to find pairs of digits from {1, 2, 3, 4, 5, 6} that can serve as the last two digits and are divisible by 4.

The possible pairs are:
- 12
- 16
- 24
- 32
- 36
- 52
- 56
- 64

Thus, we have **8 valid pairs** for the last two digits.

2. **Choose the first four digits**:
After selecting the last two digits, we will have 4 remaining digits from which we need to select the first four digits.

For example, if we choose the last two digits as 12, the remaining digits are 3, 4, 5, and 6.

3. **Calculate the arrangements**:
The first four digits can be arranged in any order. The number of ways to arrange 4 digits is given by the permutation formula \( P(n, r) = \frac{n!}{(n-r)!} \). Here, since we are using all 4 remaining digits, it simplifies to \( 4! \).

Therefore, the number of arrangements for the first four digits is:
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]

4. **Combine the counts**:
Since we have 8 choices for the last two digits and 24 arrangements for the first four digits, the total number of 6-digit numbers is:
\[ \text{Total} = \text{(Number of choices for last two digits)} \times \text{(Arrangements of first four digits)} = 8 \times 24 = 192 \]

### Final Answer:
Thus, the total number of 6-digit numbers that can be formed is **192**.

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