There are 12,033,222,880 of them.
There are 16C4 = 16*15*14*13/(4*3*2*1) = 1820 combinations.
13 combinations of 3
There are 24C12 = 24*23*...*13/(12*11*...*1) = 2,704,156 combinations.
(13 x 12 x 11 x 10)/(4 x 3 x 2 x 1) = 715 of them
If they can repeat, then: 17^6=24,137,569 If they can't repeat, then: 17*16*15*14*13*12=8,910,720
1 and 11 and 21 and 31 and 42 and 12 and 22 and 32 and 43 and 13 and 23 and 33 and 44 and 14 and 24 and 34 and 416 combinations
There are 5245786 possible combinations and I am not stupid enough to try and list what they are!
You can make 6 combinations with 3 numbers. They are: 123 213 312 132 231 321 * * * * * NO! Those are permutations! In combitorials, the order does not matter so that the combination 123 is the same as the combination 132 etc. So all of the above comprise just 1 combination. With three numbers you can have 1 combination of three numbers (as discussed above), 3 combinations of 2 numbers (12, 13 and 23) 3 combinations of 1 number (1, 2 and 3) In all, with n numbers you can have 2n - 1 combinations. Or, if you allow the null combination (that consisting of no numbers) you have 2n combinations.
Only three: 12, 13 and 23. Remember that the combinations 12 and 21 are the same.
Assuming you meant how many combinations can be formed by picking 8 numbers from 56 numbers, we have:(56 * 55 * 54 * 53 * 52 * 51 * 50 * 49)/8! = (7 * 11 * 3 * 53 * 13 * 51 * 25 * 7) = 1420494075 combinations. (Also equal to 57274321104000/40320)
As a product of its prime factors: 2*5*13 = 130
15 choose 3 = 15!/(3!*(15-3)!) = 15!/(3!*12!) = (15*14*13)/(3*2) = 5*7*13 = 455