If no repetition, 6! ie 720; if repetition allowed, 66 ie 46656.
Assuming no repeated digits, lowest first, 20; in any order 120; Allowing repeated digits: 216
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
There are 840 4-digit combinations without repeating any digit in the combinations.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
There are 5,040 combinations.
There are 210 4 digit combinations and 5040 different 4 digit codes.
the answer is = first 2-digit number by using 48= 28,82 and in 3 digit is=282,228,822,822
There are twelve possible solutions using the rule you stated.
There are 15180 combinations.
Just one. unless you count 123456 different from 132456 then there are 46656 * * * * * But you cannot count 123456 as different from 132456 since it is NOT a different combination. And the question was about combinations.
if its not alphanumeric, 999999 variations
Just the one. The order of the numbers does not matter in a combination so that 123456 is the same combination as 245136.
The order of the digits in a combination does not matter. So 123 is the same as 132 or 312 etc. There are 10 combinations using just one of the digits (3 times). There are 90 combinations using 2 digits (1 once and 1 twice). There are 120 combinations using three different digit. 220 in all.
It is: 9C7 = 36
There are 38760 combinations.
If you are allowed to repeat letters (A-Z) then there are 17,576 different combinations. If you can not repeat letters then there are 15,600 combinations.
None. Extremely simple proof: To be divisible by 5 the units digit must be 5. But then it would not be divisible by 2.