There are 3*2*1 = 6 possible orders of the three digits.
But there are also
25*7 = 224
27*5 = 640
52*7 = 175
72*5 = 245
8
6, as long as they are all different
There are no three didgit numbers but there are 63 three digit numbers.
The sum is 22 times the sum of the three digits.
There are 10 digits, but for a three digit number the first number cannot be a 0. Thus: there is a choice of 9 digits for the first (and last digit which must be the same), with 10 choices of digit for the second (middle) digit, making 9 × 10 = 90 such palindromic numbers.
8
6, as long as they are all different
There are no three didgit numbers but there are 63 three digit numbers.
The sum is 22 times the sum of the three digits.
144
24 if you don't repeat any numbers.24
25=20,21,23
There are 10 digits, but for a three digit number the first number cannot be a 0. Thus: there is a choice of 9 digits for the first (and last digit which must be the same), with 10 choices of digit for the second (middle) digit, making 9 × 10 = 90 such palindromic numbers.
30
If you want 4-digit numbers, there are 24 of them.
55----------------------------------------------------------------------------------------------From Rafaelrz.You can make, 5! = 120, five digit numbers using 1,2,3,4 and 6.
The number of six digit numbers that you can make from ten different digits ifrepetitions of same digit on the six digit number is allowed is 1 000 000 numbers(including number 000 000).If no repetitions of the the same digit are allowed then you have:10P6 = 10!/(10-6)! = 151 200 different six digit numbers(six digit permutations form 10 different digits).