A little help is needed again.?

Question

How many different 3 digit numbers can be made if the first number cannot be a 0 and the numbers can repeat?

Answer 1

Since 0 is not allowed in the first digit, then there are only 9 possibilities for the first digit.  For the second and third digits there are 10 possible numbers.  The total possible combinations then are 9 x 10 x 10 = 900.  We don't have to subtract the first hundred numbers from the total because they are automatically excluded by excluding 0 as the first digit.  We could look at all these possible numbers as a roster running from 000 to 999.  There are 1,000 numbers in this roster, but if we disallow all numbers beginning with 0, then we disallow the series 000 through 099, which includes 100 numbers.  Then the total possible number of numbers remaining is 1,000 - 100 = 900.  

We can evaluate the possibilities like this:

000 - 099  Disallowed.  There are 100 numbers in this range.

100 - 199

200 - 299

300 - 399

400 - 499

500 - 599

600 - 699

700 - 799

800 - 899

900 - 999

In each one of these ranges there are 100 numbers, not 99 as claimed by schwarzrose22.  After disallowing the first 100 numbers, we have 9 ranges with 100 numbers in each range, so the total number of possible numbers is 9 x 100 = 900, not 891 or 801.

Answer 2

801 numbers because 1-99 is not a three digit and the highest three digit is 999! Good luck!

Answer 3

3 digit numbers w/o repeating the digits. make 3 box for each digit and write how many of them is possible. first box except for 0, 9digits left.

second digit can also be 0 but since we used 1 digit for the first box again 9. the last box is 8 because we used 2 digits already. 9*9*8=648

Answer 4

891 i believe

100-199

200-299

300-399

400-499

500-599

600-699

700-799

800-899

900-999

99 * 9 = 891