How many positive integers less than 500 can be formed using the digits 2, 3,…

Concept: When items cannot repeat, the number of ordered arrangements of r positions chosen from n distinct items is the permutation nPr = n!/(n−r)!. To count numbers of different digit-lengths from a fixed digit set, split the count by length (1-digit, 2-digit, 3-digit, …) and add the disjoint cases; when a length has a value restriction (like an upper bound), first restrict the leading digit's choices, then permute the remaining positions from what's left.

Application:

1. The digit set is {2, 3, 4, 5}; any number formed uses each of its digits at most once, and no digit outside this set is allowed.

2. 1-digit numbers: any of the 4 digits can stand alone — 4 numbers, and every one of them is under 500.

3. 2-digit numbers: pick the tens digit (4 ways), then the units digit from the remaining 3 digits (3 ways): 4P2 = 4 × 3 = 12. Every such 2-digit number is under 500.

4. 3-digit numbers under 500: the hundreds digit is what decides the size, so it can only be 2, 3, or 4 (3 choices) — picking 5 there would make the number 500 or more. For each such hundreds digit, the remaining two places are a permutation of the leftover 3 digits: 3P2 = 3 × 2 = 6 arrangements. So the 3-digit count is 3 × 6 = 18.

5. Since a number has exactly one digit-length, the cases are disjoint — add them (sum/additive principle): 4 + 12 + 18.

Cross-check: count the 3-digit case a different way. Ignoring the size restriction, all 3-digit arrangements would be 4P3 = 4 × 3 × 2 = 24. Exactly the arrangements with hundreds digit 5 must be excluded: fixing 5 in front leaves 3P2 = 6 ways for the last two places, so 24 − 6 = 18 — the same count reached independently in the Application step above.

Total = 4 + 12 + 18 = 34.