111, 121, 131, 222, 212, 232, 333, 313, 323
111, 121, 222, 212
111, 121, 222, 212
111, 121, 212, 222
There are two possible digits for the first and last digit, and two possible digits for the centre digit, making 2 × 2 = 4 possible 3 digit palindromes from the set {1, 2}, namely the set {111, 121, 212, 222}.
111, 121, 131, 222, 212, 232, 333, 313, 323
111, 121, 222, 212
111, 121, 222, 212
111, 121, 212, 222
There are two possible digits for the first and last digit, and two possible digits for the centre digit, making 2 × 2 = 4 possible 3 digit palindromes from the set {1, 2}, namely the set {111, 121, 212, 222}.
111, 121, 131, 222, 212, 232, 333, 313, 323
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.
676 of them.
Yes. But that is true only if the 100 digits do not include 0. Or, if 0 is included, then you consider "0n0" to be a three digit number. Most people would consider is to be a 2-digit number.
9+3-1
it would be 321 312 231 213 123 132
The maximum number of different three letter palindromes that can be made using any letter of the alphabet is 1 - that letter repeated 3 times. However, if you meant you can pick any letter for each letter of the palindrome, then: The first and last letters are the same and there is a choice of 26 letters for them; For each choice for the first (and last) letter, there is a choice of 26 letters for the second letter Making a total of 26 × 26 = 676 such three letter palindromes (where case of the letter does not matter).
20From Rafaelrz,The number of three digit permutations you can get from the four different digits3. 6. 7. 9, is given by:4P3 = 4!/(4-3)! = 4x3x2x1/1 = 24