4
3,124,550 possible combinations
If the question is about 4 things out of 4, the answer is, obviously, 1. If the question is about 4 things out of n, then the answer is nC4 = n!/[(n-4)!*4!], whose value will obviously depend on n.
Think of either using or not using the numbers 1 and 4 for each digit. For example, let ____ ____ _____ ____ be the 4 digit number, now 1 can either go in to the first spot or 4 can. So there are 2 choices for that spot, and similarly two for the next spot and the next and the last. So there are 2^4 =16 combinations of 1 and 4 if we make 4 digit numbers. (this uses the multiplication rule) To help a little more to see it, look at just the two digit number, It should have 2^2=4 choices by the logic above. Here they are: 11,14,41,44 and that's it!. See how we had two choices for each digit?
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.
64
Not repeating, it is 7*6*5*4 which is 840 ways ---- There are 7 choices for each of four digits, right? 74 = 2401
Using unnatural color combinations. for apex :)
There are 10 possible digits that can be used: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. If you can only use each digit once, there are 10 choices for the first digit, 9 remaining choices for the second digit, and 8 remaining choices for the final digit - 10 x 9 x 8 = 720 combinations. If you can use each digit repeatedly, you have 10 choices for each digit - 10 x 10 x 10 = 1000 combinations.
11
140 possible combinations
There are 125970 combinations and I am not stupid enough to try and list them!
9
462 combinations.
There are 5,040 combinations.
3,124,550 possible combinations
126 combinations. Remember that the order does not matter.
There are 252 combinations.