Let's solve it a step at a time.
For the first digit, how many choices do you have? 9
You can choose 1..9 but not 0, so that's nine choices for the most significant digit.
For the second digit, how many choices do you have? 10
It can be 0..9.
For the third digit, you also have 10 choices.
Choosing one digit doesn't limit your choices for other digits and mirrored numbers (e.g. 123 and 321) are different, so all choices make a unique number.
So the total is the product of our three choices: 9x10x10
There are 76 such numbers. Eight more if you allow numbers to start with 0.
There are 2 possible digits for the first digit (3 or 4), leaving 3 possible digits for the second digit (5 and 6 and whichever was not chosen for the first), leaving 2 possible digits for the third. Thus there are 2 × 3 × 2 = 12 possible 3 digit numbers.
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
9x8x7x6x5x4x3x2x1 or 9! which equals 362880 possible combinations if no digits are repeated
Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.
There are 76 such numbers. Eight more if you allow numbers to start with 0.
0123456789
Assuming that the first digit of the 4 digit number cannot be 0, then there are 9 possible digits for the first of the four. Also assuming that each digit does not need to be unique, then the next three digits of the four can have 10 possible for each. That results in 9x10x10x10 = 9000 possible 4 digit numbers. If, however, you can not use the same number twice in completing the 4 digit number, and the first digit cannot be 0, then the result is 9x9x8x7 = 4536 possible 4 digit numbers. If the 4 digit number can start with 0, then there are 10,000 possible 4 digit numbers. If the 4 digit number can start with 0, and you cannot use any number twice, then the result is 10x9x8x7 = 5040 possilbe 4 digit numbers.
id say, 0123456789. but depends how many digits you can have
The number of six digit numbers that you can make from ten different digits ifrepetitions of same digit on the six digit number is allowed is 1 000 000 numbers(including number 000 000).If no repetitions of the the same digit are allowed then you have:10P6 = 10!/(10-6)! = 151 200 different six digit numbers(six digit permutations form 10 different digits).
If the numbers are not repeated then the highest 5 digit even number to be made out of 0123456789 is : 98764.
There are 4 possible numbers if the digits are not repeated; 18 if they are. Those are 3-digit numbers, assuming that zero would not be a leading digit. If zero is allowed for a leading digit, then you can have 6 for the non repeated, and 27 if repetition.
There are 9 possible numbers for the first digit (one of {1, 2, ..., 9}); with 9 possible digits for the second digit (one of {0, 1, 2, ..., 9} which is not the first digit)); with 8 possible digits for the third digit (one of {0, 1, 2, ..., 9} less the 2 digits already chosen); This there are 9 × 9 × 8 = 648 such numbers.
There are 2 possible digits for the first digit (3 or 4), leaving 3 possible digits for the second digit (5 and 6 and whichever was not chosen for the first), leaving 2 possible digits for the third. Thus there are 2 × 3 × 2 = 12 possible 3 digit numbers.
It is possible to create a 3-digit number, without repeated digits so the probability is 1.
It can have 4 digits, because the highest possible two digit numbers 99*99=9801.
There are 2941 4-digit numbers such no two of its digits differ by 1.