There are: 9C6 = 84 combinations
If the numbers are allowed to repeat, then there are six to the fourth power possible combinations, or 1296. If they are not allowed to repeat then there are only 360 combinations.
8C4 = 70
There are 32C3 = 32*31*30/(3*2*1) = 4960 combinations. I do not have the inclination to list them all.
You would get 4!/2! = 12 combinations.
The answer is 10C4 = 10!/[4!*6!] = 210
There are 840 4-digit combinations without repeating any digit in the combinations.
There are 35C4 = 35*34*33*32/(4*3*2*1) = 52,360 combinations.
Including the leading zero, then the answer is 210. Or, allowing for repeating numbers, 10000.
There are: 9C6 = 84 combinations
11
There are 167960 combinations.
It depends on combinations with how many numbers. There is only one combination of 99 numbers taken from 99.
There are 125970 combinations and I am not stupid enough to try and list them!
Not repeating, it is 7*6*5*4 which is 840 ways ---- There are 7 choices for each of four digits, right? 74 = 2401
There are 10,737,573 combinations and you must think I am mad if you think I am going to list them!
There are a huge number of combinations of 5 numbers when using the numbers 0 through 10. There are 10 to the 5th power combinations of these numbers.