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.

not only in mathematics, scientific proof also are required in the other field of science. also maths proofs are of particular structure, containing steps and logics. so proving is like doing science, it is science. that's why it is important to prove. 

an inaccurate proof leads to errors, or stating results that are not accurate.  which wil mess the probelm up you dont want this