How many 3-digit numbers can be formed using only the digits 2, 4, 6 and 8, if the repetition of digits is allowed?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 64
The three places are filled one after another and the numbers of choices are multiplied. With repetition allowed, every place can take any of the given digits. None of the digits is 0, so the first place has no restriction.
Hundreds place: 4 choices (2, 4, 6 or 8)
Tens place: 4 choices (a digit may be used again)
Units place: 4 choices
Numbers \(=4\times4\times4=64\)
Why the other options are incorrect
Option (1) 24 — Incorrect. This is \(4\times3\times2\), the count when repetition is not allowed.
Option (3) 12 — Incorrect. This is \(4\times3\), which fills only two places.
Option (4) 81 — Incorrect. This is \(3^4\), with the roles of digits and places swapped; it should be 4 choices for each of 3 places, \(4^3\).