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.
There is only one combination. The order of the digits in combinations makes no difference. They are considered as being different if they are permutations, not combinations.
There are only 10 combinations. In each combination one of the 10 digits is left out.
Since the order of the digits does not matter there are only five combinations: 1234, 1235, 1245, 1345 and 2345.
Assuming no repeated digits, lowest first, 20; in any order 120; Allowing repeated digits: 216
6 for 3-digits, 6 for 2-digits, 3 for 1-digits, and 15 for all of the combinations
If you use them only once each, you can make 15 combinations. 1 with all four digits, 4 with 3 digits, 6 with 2 digits, and 4 with 1 digit. There is also a combination containing no digits making 16 = 24 combinations from 4 elements.
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
5040, assuming none of the digits are the same. (Assuming they're not, there's 5040 unique combinations you can make out of 7 digits).
Oh, dude, you're hitting me with some math vibes here. So, if you have 6 digits to choose from to make a 4-digit combination, you can calculate that by using the formula for permutations: 6P4, which equals 360. So, like, you can make 360 different 4-digit combinations from those 6 digits. Math is wild, man.
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.
9.
There is only one combination. The order of the digits in combinations makes no difference. They are considered as being different if they are permutations, not combinations.
∞ \ Infinite
There are infinite combinations that can make 3879
1
32C3 = 4960