90 of them.
You can select 9 numbers for the first digit, 8 numbers for the second digit, and 7numbers for the third digit; so 504 (e.g. 9*8*7) different three digit numbers can be written using the digits 1 through 9.
24
With 123 digits you can make 123 one-digit numbers.
a 8-digits palindromic number is a number consisting of a 4-digits number written and then written backward, i.e. 1234 4321 so there are as many palindromic 8-digits numbers as 4-digits numbers so from 1000 to 9999, so there are 8999 palindromic 8 digits number (I assumed that 00011000 is not to be considered as a valid 8 digit number)
This cannot be answered. This will have to be written to where it can be understood.
There are 5460 five digit numbers with a digit sum of 22.
5040 different 4 digit numbers can be formed with the digits 123456789. This is assuming that no digits are repeated with each combination.
There are 9 digits that can be the first digit (1-9); for each of these there is 1 digit that can be the second digit (6); for each of these there are 10 digits that can be the third digit (0-9); for each of these there are 10 digits that can be the fourth digit (0-9). → number of numbers is 9 × 1 × 10 × 10 = 900 such numbers.
27 three digit numbers from the digits 3, 5, 7 including repetitions.
They are written as numbers usually are. The place value of the digit immediately to the left of the decimal point is ones and the place value of all other digits is ten times the value of the digit to their right.
To find the even two-digit numbers where the sum of the digits is 5, we need to consider the possible combinations of digits. The digits that sum up to 5 are (1,4) and (2,3). For the numbers to be even, the units digit must be 4, so the possible numbers are 14 and 34. Therefore, there are 2 even two-digit numbers where the sum of the digits is 5.
Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.