Question

A student is asked to form numbers between 3000 and 9000 with digits 2, 3, 5, 7 and 9. If no digit is to be repeated, in how many ways can the student do so?

• A. 24
• B. 120
• C. 60
• D. 72

Hint:

Fix the most significant digit to restrict range, then count permutations of remaining digits.

Answer 1

The Correct Option is C

We need 4-digit numbers between 3000 and 9000 using the digits 2, 3, 5, 7, 9 with no repetition.
The number must be 4-digit, and must be \( \geq 3000 \) and \(<9000 \)
So the thousands place can be 3, 5, 7 (since 2<3000 and 9 makes 9000 or more)
Valid leading digits = 3, 5, 7 → 3 choices
Remaining digits from the set of 5, without repetition. So for each choice: \[ \text{Ways} = 3 \times 4 \times 3 \times 2 = \boxed{72} \] BUT we must also ensure that number<9000 → So leading digit = 3, 5, 7 only
If leading digit is 9 → number ≥ 9000 → Not valid
Check total:
From digits: 2, 3, 5, 7, 9 → Total 5 digits
Select 4 digits out of them → \( {}^5P_4 = 5 \cdot 4 \cdot 3 \cdot 2 = 120 \)
Now remove those starting with 9 → Remaining = 120 − 24 = \boxed{96}
Wait, only numbers between 3000 and 9000 allowed → only those with 1st digit 3, 5, or 7 Try explicitly:
First digit choices = 3 (available), 5, 7 → 3 choices For each: choose 3 digits from remaining 4 → \( {}^4P_3 = 4 \cdot 3 \cdot 2 = 24 \) Total = \( 3 \cdot 24 = \boxed{72} \)