I'll assume that you mean " ... from the digits 0, 1, 2, 3, 4, 5, and 6 ?" .
-- If the same digit can appear more than once in the same combination,
then there are (7 x 7 x 7) = 343 possibilities.
-- If the three digits in one combination must all be different digits,
then there are (7 x 6 x 5) = 210 possibilities.
-- If the combinations are actual cou nting numbers, and so the first digit
can't be 'zero', then there are (6 x 6 x 5) = 180 possibilities.
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about 1,0000000000000
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.
10C6 = 10*9*8*7/(4*3*2*1) = 210 combinations.
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.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.