Q.How many 3 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated?
Count how many 3-digit even numbers can be built from the digit set when digits may repeat, by fixing the units digit to be even first and then filling the other two places freely.
Multiplication (Fundamental) Principle of Counting: if a task consists of independent stages with choices respectively, the total number of ways to complete the whole task is . For a number to be even, its last (units) digit must itself be even — this constraint is handled first since it restricts the available digit pool for that one position.
- Units place (rightmost digit) — a number is even exactly when its last digit is even. From the digit set , the even digits are and only, giving 2 choices for the units place.
- Hundreds place (leftmost digit) — since repetition of digits is allowed, and every digit in is non-zero (no leading-zero issue to worry about), any of the 5 digits may be used here, independent of what was chosen for the units place: 5 choices.
- Tens place (middle digit) — again, repetition is allowed and there is no restriction on this digit, so any of the 5 digits may be used: 5 choices.
- By the multiplication principle, since the three digit-choices are made independently (choosing the units digit doesn't reduce the pool for hundreds/tens, because repetition is permitted), multiply the choice-counts: .
- Compute: , then .
Self-check: List-style sanity check — for each of the 2 valid units digits (2 or 4), there are ways to fill the other two places, giving total, matching the multiplication-principle result. ✓
three-digit even numbers
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