Question

The total number of 7 digit positive integral numbers with distinct digits that can be formed using the digits 4, 3, 7, 2, 1, 0, 5 is:

• A. \( 4320 \)
• B. \( 4340 \)
• C. \( 4310 \)
• D. \( 4230 \)
• E. \( 4220 \)

Hint:

When 0 is included in your set of digits and you need to form an \( n \)-digit number using all \( n \) digits, the formula is always \( (n-1) \times (n-1)! \). Here, \( (7-1) \times 6! = 6 \times 720 = 4320 \).

Answer 1

The Correct Option is A

Concept: To form a 7-digit number, we have 7 available positions. The most important rule is that the first digit (million's place) cannot be zero, otherwise it becomes a 6-digit number.

Step 1: Analyze the available digits.
The set of digits provided is: \( \{0, 1, 2, 3, 4, 5, 7\} \). There are exactly 7 digits in total.

Step 2: Fill the positions one by one.

• 1st Position (leftmost): Can be any digit except 0. Options = 6 (1, 2, 3, 4, 5, or 7).
• 2nd Position: Can be any of the remaining 6 digits (including 0). Options = 6.
• 3rd Position: Remaining digits. Options = 5.
• 4th Position: Remaining digits. Options = 4.
• 5th Position: Remaining digits. Options = 3.
• 6th Position: Remaining digits. Options = 2.
• 7th Position: Last remaining digit. Options = 1.

Step 3: Apply the Multiplication Principle.
Total numbers = \( 6 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) \[ \text{Total} = 6 \times (6!) \] \[ \text{Total} = 6 \times 720 \] \[ \text{Total} = 4320 \]