How many combinations of three digits can be made from the numbers 0-9?

Answer 1

9

Answer 2

The question does NOT say that the three places must be different digits, so repetition is allowed. The first digit can't be zero, so there are 9 choices for it.

The second digit can be any one of the 10, including zero.

The third digit can also be any one of the 10, including zero. So the quantity of 3-digit numbers that can be written is (9 x 10 x 10) = 900. (This makes sense. The smallest number you can write with 3 digits is 100, and the largest is 999, which is a total of 900 different numbers.)

If repetition is not allowed, then: The first digit can't be zero, so there are 9 choices for it.

For each choice, the second digit can be any one of the remaining 9 (including zero).

For each choice, the third digit can be any one of the remaining 8 (including zero). So the quantity of 3-digit numbers that can be written without repetition or a leading zero is (9 x 9 x 8 ) = 648.

But the question doesn't ask "how many numbers can be made".

It asks "How many combinations can be made ?" When you say "combinations" in math, that ordinarily means that the order doesn't matter, i.e. 247 would be considered the same thing as 724. Granted, this is getting pretty picky, but that's exactly what math is all about, so we need to consider the possibility that the question was actually referring to combinations, and the order of the digits doesn't matter. If repetition is not allowed: The first digit can be any one of 10, including zero.

For each choice, the second digit can be any one of the remaining 9.

For each choice, the third digit can be any one of the remaining 8.

The total number of 3-digit groups is (10 x 9 x 8) = 720.

Each group of 3 digits can be arranged in (3 x 2) = 6 ways. So if the order doesn't matter, there are (720/6) = 120separate, distinct COMBINATIONS of 3 digits.