Solve the problem.
-How many 3-digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, if repetition of digits is allowed?

Question

Quizplus · Accepted Answer

For a 3-digit number, the first digit can be any of the 9 options (1-9, since it can't be 0), and the second and third digits can be any of the 10 digits (0-9), allowing repetition. Thus, the total number of combinations is 9 * 10 * 10 = 900. However, the question asks for the number of 3-digit numbers that can be formed, which includes all numbers from 100 to 999. Since 999 is the last 3-digit number, subtracting 100 (the first 3-digit number) from 999 gives 899, but since both 100 and 999 are inclusive, we add 1 back, resulting in 900 total numbers, which is not listed as an option. The correct interpretation should be calculating the total possibilities directly, which is indeed 900, not 1000. My explanation mistakenly suggested adding an extra step, but the correct calculation directly yields 900 possibilities for forming 3-digit numbers with the given conditions.

Solve the Problem. -How Many 3-Digit Numbers Can Be Formed Using the Digits