How many positive integers less than 700 can be formed using the digits 2, 4,…

Concept: To count integers below a given bound that can be formed from a fixed set of digits without repetition, split the count by number of digits used. For each digit-length k, the number of valid arrangements is a permutation ⁿPₖ = n × (n−1) × … × (n−k+1); when the bound restricts which digit can occupy the leading place, only the allowed leading digits are counted for that digit-length, and the totals across digit-lengths are added.

Working:

1. One-digit numbers: any of the 4 digits {2, 4, 6, 7} standing alone is a valid 1-digit number, and every 1-digit number is automatically less than 700. Count = 4P1 = 4.

2. Two-digit numbers: choose an ordered pair of distinct digits from the 4 available digits (each digit used at most once). Count = 4P2 = 4 × 3 = 12. Every 2-digit number formed this way is well below 700, so no extra restriction applies here.

3. Three-digit numbers: here the bound matters — a 3-digit number is below 700 only if its hundreds digit is less than 7, so the hundreds place can only be 2, 4, or 6 (3 choices). Once the hundreds digit is fixed, the remaining two places are filled by an ordered arrangement of the 3 leftover digits: 3P2 = 3 × 2 = 6. Count = 3 × 6 = 18.

4. Add the three cases: 4 + 12 + 18 = 34.

Cross-check: Ignoring the bound, all 3-digit arrangements of the 4 digits number 4P3 = 4 × 3 × 2 = 24. Exactly those with 7 in the hundreds place (1 choice for hundreds × 3P2 = 6 arrangements for the rest = 6 numbers) exceed the bound, leaving 24 − 6 = 18 valid 3-digit numbers — matching the count above, so the total of 34 stands confirmed.