Compute the permutations and combinations.

How many three-digit numbers greater than 200 can be formed from the digits 1, 2, 6, 7, and 9, if the digits can be repeated?
I need you to show your work,

100 numbers

Step-by-step explanation:

We only want three-digit numbers, so let's provide three spaces:

__ __ __

The first number can be 2, 6, 7, or 9; it can't be 1 because then we're going to have a number in the 100s, which will be less than 200 (and that's not allowed). So, there are 4 ways to choose the first number.

The second number, though, can be any of the 5 numbers because once we've established the first number, the second digit is anything. So there are 5 ways to choose this one.

Finally, there are also 5 digits to choose from for the last digit because we can reuse digits.

Multiply these together:

4 * 5 * 5 = 100

There are 100 three-digit numbers greater than 200 that can be formed with the given digits.

2

step-by-step explanation: