I am assuming you mean 3-number combinations rather than 3 digit combinations. Otherwise you have to treat 21 as a 2-digit number and equate it to 1-and-2. There are 21C3 combinations = 21*20*19/(3*2*1) = 7980 combinations.
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
There are 9C2 = 9*8/(2*1) = 36 2-digit combinations.
If the same 7 digits are used for all the combinations then n! = 7! = 7*6*5*4*3*2*1 = 5040 combinations There are 9,999,999-1,000,000+1=9,000,000 7-digit numbers.
There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.
How many four digit combinations can be made from the number nine? Example, 1+1+2+5=9.
There are different numbers of combinations for groups of different sizes out of 9: 1 combination of 9 digits 9 combinations of 1 digit and of 8 digits 36 combinations of 2 digits and of 7 digits 84 combinations of 3 digits and of 6 digits 126 combinations of 4 digits and of 5 digits 255 combinations in all.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
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.
The question is incorrect since there are 210 - 1 possible combinations. The digit 0 can be in the combination or out. That gives 2 ways. With each, the digit 1 can be in the combination or out - 2*2 = 22 ways. With each, the digit 2 can be in the combination or out = 23 ways. With each, the digit 3 can be in the combination or out = 24 ways. etc. So 210 ways in all except that one of them is the null combination. Now 210 = 1024 so there are only 1023 combinations. If, instead, you allow the digits to be used many times, there is no limit to the number of combinations.
the whole numbers 2000 to 2999
There are 28706 such combinations. 5456 of these comprise three 2-digit numbers, 19008 comprise two 2-digit numbers and two 1-digit numbers, 4158 comprise one 2-digit number and four 1-digit numbers and 84 comprise six 1-digit numbers.