Question

If all the numbers which are greater than 6000 and less than 10000 are formed with the digits \( 3,5,6,7,8 \) without repetition of the digits, then the difference between the number of odd numbers and the number of even numbers among them is:

• A. \( ^{4}P_3 \)
• B. \( 3(^{4}P_2) \)
• C. \( ^{5}P_3 \)
• D. \( 2(^{4}P_3) \)

Hint:

To count numbers within a range, first determine valid leading digits. Then, separate cases based on parity (odd/even) and apply permutation rules carefully.

Answer 1

The Correct Option is A

We are given the digits \( \{3, 5, 6, 7, 8\} \) and need to form 4-digit numbers between 6000 and 9999 without repetition. 

Step 1: Identify Valid Leading Digits 
- A number must be \(\geq 6000\), so the first digit must be 6, 7, or 8 (valid leading digits). - The remaining 3 digits are selected from the remaining 4 digits. 

Step 2: Count Even and Odd Numbers 
- A number is even if its last digit is even (\(6,8\)). - A number is odd if its last digit is odd (\(3,5,7\)). 

Step 3: Count Odd and Even Cases 
1. Case 1: Last digit is even (6 or 8) - First digit choices: \( 6, 7, 8 \) (3 choices). - Last digit choices: \( 6 \) or \( 8 \) (2 choices). - Middle two digits are chosen from remaining 3 digits: \( ^3P_2 \). \[ \text{Total even numbers} = 3 \times 2 \times {}^3P_2. \] 2. Case 2: Last digit is odd (3,5,7) - First digit choices: \( 6, 7, 8 \) (3 choices). - Last digit choices: \( 3,5,7 \) (3 choices). - Middle two digits are chosen from remaining 3 digits: \( ^3P_2 \). \[ \text{Total odd numbers} = 3 \times 3 \times {}^3P_2. \] 

Step 4: Compute Difference Between Odd and Even Numbers 
\[ \text{Difference} = (3 \times 3 \times {}^3P_2) - (3 \times 2 \times {}^3P_2). \] \[ = 3(3 - 2) {}^3P_2 = 3 {}^3P_2 = {}^4P_3. \] 

Step 5: Conclusion 
Thus, the difference between the number of odd and even numbers is: \[ \boxed{^{4}P_3}. \]