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?
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Respuesta :

Answer:

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.

Answer:

100

Step-by-step explanation:

100

Step-by-step explanation:

If three-digit numbers greater than 200 can be formed from the digits 1, 2, 6, 7, and 9

Here we have total number of digits are 5

We need to make 3-digit number which is greater than 200

First digit must be 2 or greater than 2

From list of number 1,2,6,7,9

Four numbers are greater than equal to 2

For 1st place total number of possibility = 4

For 2nd place number of possibility = 5

For 3rd place number of possibility = 5

Total numbers of which are greater than 200 = 4 x 5 x 5 = 100

Hence, 100 three-digit number greater than 200

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