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Question
How many three-digit numbers can be formed from the digits 0, 1, 3, 5, 6 if repetitions of digits are not allowed?
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Solution
Step 1: Choose the hundreds digit (first digit)
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Cannot be 0.
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From {1, 3, 5, 6} → 4 options
Step 2: Choose the tens digit (second digit)
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One digit is already used in hundreds place.
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Remaining digits = 4
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But includes 0 (which is now allowed) → So 4 options
Step 3: Choose the units digit (third digit)
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Two digits already used.
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Remaining digits = 3 → So 3 options
Total numbers = 4 (choices for first) × 4 (second) × 3 (third)
= 48
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