Assuming no repetition, 12 x 11 x 10.... x 4 x 3 x 2.
If repetition is allowed, 1212
There are 24C12 = 24!/[12!*12!] = 2,704,156 combinations.
12
2,704,156
You would get 4!/2! = 12 combinations.
There are 24C12 = 24*23*...*13/(12*11*...*1) = 2,704,156 combinations.
12
30
Only three: 12, 13 and 23. Remember that the combinations 12 and 21 are the same.
26 = 64 combinations, including the null combination - which contains no numbers.
Assuming that the only permitted operation is addition, then there are 4 combinations.
11
There are 12C4 4 NUMBER combinations. And that equals 12*11*10*9/(4/3/2/1) = 495 combinations. However, some of these, although 4 number combinations consist of 7 digits eg 1, 10, 11, and 12. Are you really sure you want 4-DIGIT combinations?