Wiki User
∙ 6y ago111, 121, 222, 212
Wiki User
∙ 6y ago111, 121, 222, 212
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, 212, 222
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.
111, 121, 131, 222, 212, 232, 333, 313, 323
111, 121, 222, 212
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, 212, 222
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.
111, 121, 131, 222, 212, 232, 333, 313, 323
Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.
None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits.
24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.
9+3-1
There are 900 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.
972.