How many 5-digit positive integers can be formed using the digits 3, 4, and 5, if each digit must appear at least once?

Mon, 04 May 2026 21:53:54 +0000

**How Many 5-Digit Positive Integers Can Be Formed Using the Digits 3, 4, and 5—If Each Digit Must Appear at Least Once?** When you imagine all possible 5-digit numbers built from just 3, 4, and 5, the realm of combinations feels expansive—yet tightly constrained by one simple rule: every digit 3, 4, and 5 must show up at least once. So, how many such unique numbers exist? It’s a puzzle where variation meets inclusion, creating a surprisingly grounded exercise in combinatorics. At first glance, forming a 5-digit number with exactly five digits using three choices per position seems straightforward. Without restrictions, each digit can be 3, 4, or 5—giving 3⁵ = 243 total combinations. But this count includes many invalid ones: numbers missing digits, using only 3s and 4s, or skipping 5 entirely. In reality, only those sequences where all three digits appear at least once qualify. To find the valid count, mathematicians use the principle of inclusion-exclusion—a method designed to subtract invalid cases from the total. First, subtract all 5-digit strings missing at least one digit. There are three cases: numbers missing 3, missing 4, and missing 5. Each missing one digit restricts choices to two digits, producing 2⁵ = 32 options per exclusion. Three such exclusions give 3 × 32 = 96 invalid sequences.