None. The only way for it to be possible would be 3 zeros which is not considered a 3 digit numbers.
27 three digit numbers from the digits 3, 5, 7 including repetitions.
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.
24
There are 5*4*3 = 60 such numbers.
There are several ways to answer this. I think this is the easiest: Count the number of three-digit numbers: The first digit can be 1-9; the second and third digits can be 0-9. So, there are 9 choices for the first digit and 10 choices for the second and third digits. This gives you 9 x 10 x 10 = 900 three-digit numbers. Right? Then count the number of three-digit numbers that do NOT contain the digit 2. This is done similarly! Try it! Now, if you think about it, the difference between these two quantities IS the number of three-digit numbers that DO contain a 2, which is what you wanted.
There are 900 three-digit numbers.
757
27 three digit numbers from the digits 3, 5, 7 including repetitions.
21
874
21.
450.
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.
1
There are 900 of them.
With 123 digits you can make 123 one-digit numbers.
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.