How many 3-digit numbers can be formed using the digits 0, 2, 4, 6 and 8, if the repetition of digits is allowed?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 100
A 3-digit number cannot begin with 0. With repetition allowed, the other places can take any of the digits.
Hundreds place: 2, 4, 6 or 8, that is 4 choices
Tens place: 5 choices; units place: 5 choices
Numbers \(=4\times5\times5=100\)
Why the other options are incorrect
Option (1) 125 — Incorrect. This is \(5^3\), which lets a number begin with 0.
Option (3) 48 — Incorrect. This is \(4\times4\times3\), the count without repetition.
Option (4) 80 — Incorrect. This is \(4\times5\times4\); with repetition the units place also has all 5 digits.