The number of four-digit numbers strictly greater than 4321 that can be formed using the digits 0, 1, 2, 3, 4, 5 (repetition of digits is allowed) is
- A.306
- B.310
- C.360
- D.288
Show answer and solution
Correct option: B
Solution
Fix thousands digit cases:
Thousands digit = 5: Any combination for remaining 3 digits → 6³ = 216 numbers.
Thousands digit = 4: Need numbers > 4321 with thousands digit 4.
- Hundreds digit = 4 or 5 (i.e., > 3): hundreds digit ∈ {4,5} → 2 × 6² = 72 numbers (all > 4300 > 4321... wait, need to check 4300s)
Let me be precise. Need 4XYZ > 4321:
- Hundreds digit > 3: digit ∈ {4,5} → 2 × 6 × 6 = 72
- Hundreds digit = 3, tens digit > 2: digit ∈ {3,4,5} → 3 × 6 = 18
- Hundreds digit = 3, tens digit = 2, units digit > 1: digits ∈ {2,3,4,5} → 4
Total with thousands digit 4: 72 + 18 + 4 = 94
Grand total: 216 + 94 = 310
The answer is option 2 (310).
Thousands digit = 5: Any combination for remaining 3 digits → 6³ = 216 numbers.
Thousands digit = 4: Need numbers > 4321 with thousands digit 4.
- Hundreds digit = 4 or 5 (i.e., > 3): hundreds digit ∈ {4,5} → 2 × 6² = 72 numbers (all > 4300 > 4321... wait, need to check 4300s)
Let me be precise. Need 4XYZ > 4321:
- Hundreds digit > 3: digit ∈ {4,5} → 2 × 6 × 6 = 72
- Hundreds digit = 3, tens digit > 2: digit ∈ {3,4,5} → 3 × 6 = 18
- Hundreds digit = 3, tens digit = 2, units digit > 1: digits ∈ {2,3,4,5} → 4
Total with thousands digit 4: 72 + 18 + 4 = 94
Grand total: 216 + 94 = 310
The answer is option 2 (310).
[JEE Main 2019 · 8 Apr · Shift II]