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Question
How many three-digit numbers are there, with no digit repeated?
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Solution
Total number of 3-digit numbers = Number of arrangements of 10 numbers, taken 3 at a time = 10P3 =\[\frac{10!}{7!} = 10 \times 9 \times 8 = 720\]
Total number of 3-digit numbers, having 0 at its hundred's place = 9P2 =\[\frac{9!}{7!} = 9 \times 8 = 72\]
Total number of 3-digit numbers with distinct digits = 10P3\[-\] 9P2 = 720\[-\] 72 = 648
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