To determine how many school ID numbers can be created with digits that can be repeated, where the first digit cannot be 0, we can analyze the situation as follows: The first digit can be any of the digits from 1 to 9 (9 options), while the remaining digits (assuming a total of n digits) can be any digit from 0 to 9 (10 options each). Therefore, the total number of ID numbers can be calculated as (9 \times 10^{(n-1)}), where (n) is the total number of digits in the ID.
Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.
987,653
500
It is 120 if the digits cannot be repeated.
There are only two smaller 3-digit numbers and both of them have repeated digits.
It is possible to create a 3-digit number, without repeated digits so the probability is 1.
757
Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.
987,653
9x9x8x7x6x5x4x3 =1632960
27 three digit numbers from the digits 3, 5, 7 including repetitions.
24
500
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.
1 set
It is 120 if the digits cannot be repeated.
There are only two smaller 3-digit numbers and both of them have repeated digits.