Probability that a random 5-digit number formed with digits 1–5 (repetition allowed) contains all digits at least once.

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Total 5-digit numbers = 5^5 = 3,125 Numbers missing at least one digit use inclusion-exclusion: - Missing 1 digit: C(5,1)×4^5 = 5×1,024 = 5,120 - Missing 2 digits: C(5,2)×3^5 = 10×243 = 2,430 - Missing 3 digits: C(5,3)×2^5 = 10×32 = 320 - Missing 4 digits: C(5,4)×1^5 = 5×1 = 5 By inclusion-exclusion: 5,120 2,430+320 5 = 3,005 numbers missing 1 digit Numbers containing all digits = 3,125 3,005 = 120 Probability = 120 / 3,125 = 24/625

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Asked 12 days ago|2/23/2026 9:23:38 AM

Updated 12 days ago|2/23/2026 10:00:21 AM

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cruz30

Total 5-digit numbers = 5^5 = 3,125
Numbers missing at least one digit use inclusion-exclusion:
- Missing 1 digit: C(5,1)×4^5 = 5×1,024 = 5,120
- Missing 2 digits: C(5,2)×3^5 = 10×243 = 2,430
- Missing 3 digits: C(5,3)×2^5 = 10×32 = 320
- Missing 4 digits: C(5,4)×1^5 = 5×1 = 5
By inclusion-exclusion: 5,120 2,430+320 5 = 3,005 numbers missing 1 digit
Numbers containing all digits = 3,125 3,005 = 120
Probability = 120 / 3,125 = 24/625

Added 12 days ago|2/23/2026 9:59:47 AM

This answer has been confirmed as correct and helpful.

Salas1987

5^5 - C(5,1)4^5 + C(5,2)3^5 - C(5,3)2^5 + C(5,4)1^5 - C(5,5)0^5 = 3125 - 51024 + 10243 - 1032 + 51 - 10 = 3125 - 5120 + 2430 - 320 + 5 = 120.
The probability is 120 / 3125 = 24 / 625.

Added 12 days ago|2/23/2026 10:00:21 AM

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