Use the permutation (no repetition) formula: n! / (n-r)!, where ! is the factorial. Factorial of n = n * (n-1) * (n-2) * ...* 1.
n = 4, because there are 4 items. r = 3 because you take them 3 at a time.
4! / (4-3)! = 4! / 1! = 4 * 3 * 2 * 1 = 24
64 different whole numbers can be written with 6 bits.
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.
60
89,999 different numbers i guess
There are only five distinct odd digits.
64 different whole numbers can be written with 6 bits.
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.
60
Spell_out_whole_numbers
89,999 different numbers i guess
There are 7,290 different 4-digit numbers that can be formed from the digits 1-9 without repetition.
There are only five distinct odd digits.
Different schemes have different id numbers of different lengths.
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)
Assuming you mean permutations of three digits, then the set of numbers that can be made with these digits is: 345 354 435 453 534 543 There are six possible permutations of three numbers.
12689 14689 12489
This cannot be answered. This will have to be written to where it can be understood.