Question

Let \( A = \{0,2,4,6,8\} \). The number of 5-digit numbers that can be formed using the digits in \( A \) without replacement, is

• A. \(120 \)
• B. \(96 \)
• C. \(88 \)
• D. \(64 \)
• E. \(32 \)

Hint:

For number formation, always subtract cases where leading digit is zero.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The problem asks for the number of permutations of 5 distinct digits with the restriction that the resulting number must be a 5-digit number. This implies the first digit cannot be 0.
Step 2: Key Formula or Approach:
We can solve this using two main methods: Method 1 (Indirect): Calculate the total number of permutations of the 5 digits and subtract the number of permutations that start with 0. Method 2 (Direct): Fill the positions of the 5-digit number one by one, considering the restrictions at each step.
Step 3: Detailed Explanation:
Method 1: Indirect Calculation
1. Total possible arrangements: The total number of ways to arrange the 5 distinct digits \{0, 2, 4, 6, 8\} is \(5!\). \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \] 2. Invalid arrangements (starting with 0): An arrangement starting with 0 is not a 5-digit number. If we fix the first digit as 0, we are left with arranging the remaining 4 digits \{2, 4, 6, 8\}. The number of ways to do this is \(4!\). \[ 4! = 4 \times 3 \times 2 \times 1 = 24 \] 3. Valid 5-digit numbers: The total number of valid 5-digit numbers is the total arrangements minus the invalid ones. \[ \text{Number of 5-digit numbers} = 120 - 24 = 96 \] Method 2: Direct Calculation
Consider the 5 positions of the number: \_ \_ \_ \_ \_ 1. First digit (Ten Thousands place): This digit cannot be 0. So, we can choose from \{2, 4, 6, 8\}. There are 4 choices. 2. Second digit (Thousands place): We can now use 0. Since one non-zero digit has been used, there are 4 digits remaining. So, there are 4 choices. 3. Third digit (Hundreds place): Two digits have been used. There are 3 remaining choices. 4. Fourth digit (Tens place): Three digits have been used. There are 2 remaining choices. 5. Fifth digit (Units place): Four digits have been used. There is 1 remaining choice. Total number of ways = \(4 \times 4 \times 3 \times 2 \times 1 = 96\).
Step 4: Final Answer:
The number of 5-digit numbers that can be formed is 96.