Question

Suppose we want to form three-digit numbers using the set of digits 0, 1, 2, 3, 4, 5. For example, 301 and 223 are such numbers, but 037 is not. How many such numbers are possible?

Answer 1

There are 180 three-digit numbers that can be formed using the set of digits 0, 1, 2, 3, 4, 5, considering that the first digit cannot be 0. This is calculated by multiplying the possible choices for each digit position (5 for the first digit, 6 for both the second and third digits).

Step-by-step explanation:

To find how many three-digit numbers can be formed using the set of digits 0, 1, 2, 3, 4, 5, we need to consider each digit position independently, since each has its own set of constraints.

The first digit can be anything but 0 to ensure it's a three-digit number, while the second and third digits can be any of the six possible digits.

For the first digit, we have five choices (1, 2, 3, 4, 5). The digit 0 is excluded because the number must be a three-digit number.

For the second digit, all six digits are possible (0, 1, 2, 3, 4, 5).

For the third digit, again, all six digits are possible (0, 1, 2, 3, 4, 5).

Using multiplication principle, we multiply the number of possibilities for each digit together: 5 (choices for the first digit) × 6 (choices for the second digit) × 6 (choices for the third digit).

Therefore, the total number of three-digit numbers that can be formed is 5 × 6 × 6 = 180.