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
No, 100 different digits cannot make 10,010,010,000 different three-digit palindromes. A three-digit palindrome has the form "ABA," where A is the first and last digit, and B is the middle digit. Since A can be any digit from 1 to 9 (for the first digit) and B can be any digit from 0 to 9, there are only 9 options for A and 10 options for B, resulting in a total of 90 unique three-digit palindromes (9 x 10 = 90).
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
No, 100 different digits cannot make 10,010,010,000 different three-digit palindromes. A three-digit palindrome has the form "ABA," where A is the first and last digit, and B is the middle digit. Since A can be any digit from 1 to 9 (for the first digit) and B can be any digit from 0 to 9, there are only 9 options for A and 10 options for B, resulting in a total of 90 unique three-digit palindromes (9 x 10 = 90).
Yuo can make only one combination of 30 digits using 30 digits.
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.
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.
Yes.
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