Let ' m ' denotes the number of four digit numbers such that the left most digit is odd, the second digit is even and all four digits are different and ' n ' denotes the number of four digit numbers such that left most digit is even, second digit is odd and all four digit are different. If m=nk, then k equals :

Options

A. `(4)/(5)`
B. `(3)/(4)`
C. `(5)/(4)`
D. `(4)/(3)`

Answer

To solve the problem, we need to calculate the values of \( m \) and \( n \) based on the given conditions and then find the value of \( k \) such that \( m = nk \).

### Step 1: Calculate \( m \)

1. **Determine the leftmost digit (odd)**: The leftmost digit must be odd. The odd digits available are \( 1, 3, 5, 7, 9 \). Thus, there are **5 choices** for the leftmost digit.

2. **Determine the second digit (even)**: The second digit must be even. The even digits available are \( 0, 2, 4, 6, 8 \). Thus, there are **5 choices** for the second digit.

3. **Choose the remaining two digits**: Since all four digits must be different, we have already used 2 digits. Therefore, we have 8 remaining digits to choose from (10 total digits - 2 used). We need to choose 2 digits from these 8, which can be done in \( \binom{8}{2} \) ways. After choosing, we can arrange these 2 digits in \( 2! \) ways.

Putting it all together, we have:
\[ m = 5 \times 5 \times \binom{8}{2} \times 2! \]

### Step 2: Calculate \( n \)

1. **Determine the leftmost digit (even)**: The leftmost digit must be even. The available even digits are \( 2, 4, 6, 8 \) (we cannot use 0 as it would make it a 3-digit number). Thus, there are **4 choices** for the leftmost digit.

2. **Determine the second digit (odd)**: The second digit must be odd. The available odd digits are \( 1, 3, 5, 7, 9 \). Thus, there are **5 choices** for the second digit.

3. **Choose the remaining two digits**: Similar to before, we have already used 2 digits, leaving us with 8 remaining digits to choose from. We can choose 2 digits from these 8 in \( \binom{8}{2} \) ways and arrange them in \( 2! \) ways.

Putting it all together, we have:
\[ n = 4 \times 5 \times \binom{8}{2} \times 2! \]

### Step 3: Relate \( m \) and \( n \)

We know that \( m = nk \). Substituting the expressions for \( m \) and \( n \):
\[ 5 \times 5 \times \binom{8}{2} \times 2! = (4 \times 5 \times \binom{8}{2} \times 2!) \times k \]

### Step 4: Simplify to find \( k \)

We can cancel \( \binom{8}{2} \) and \( 2! \) from both sides:
\[ 5 \times 5 = 4 \times 5 \times k \]

Dividing both sides by \( 5 \):
\[ 5 = 4k \]

Now, solving for \( k \):
\[ k = \frac{5}{4} \]

### Conclusion

Thus, the value of \( k \) is:
\[ \boxed{\frac{5}{4}} \]