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111, 121, 131, 222, 212, 232, 333, 313, 323

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7y ago

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Related Questions

What are the three digit palindromes that you can make only using the digits 1 and 2?

111, 121, 222, 212


How many three-digit palindromes can you make using only the digits 1 and 2?

111, 121, 222, 212


How many different three-digit palindromes can you make using only the digits 1 and 2?

111, 121, 212, 222


How many different three digit palindromes can you make using only the digits 1and 2?

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}.


How many different three-digit palindromes can you make using only the digits 1 2 and 3?

111, 121, 131, 222, 212, 232, 333, 313, 323


What is the maximum number of different three digit palindromes you can make using any digit?

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.


What is the maximum number of different three-letter palindromes that you can make using any letters of the alphabet that you like?

676 of them.


Can 100 different digits make 100 multiplied by 100 equal 10000 different three digits palindromes?

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.


How do you make a two digit number using three digits?

9+3-1


How many three digit numbers can you make using the one two and three?

it would be 321 312 231 213 123 132


What is the maximum number of different three letter palindromes you can make using any letter of the alphabet?

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).


How many three digit numbers can you make using 3 6 7 9 using each digit only once in every number?

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