Assuming that the digits 0-9 are available, the answer is 5040. This is because for the first digit you can pick any of the original 10 digits. For the second you can have any otherdigit, so 9 possible choices. The third follows the same rule, so has 8 possibilities, and the forth digit, 7.
This means that the total number of combinations is 10 x 9 x 8 x 7 = 5040
There are 5,461,512 such combinations.
64
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.
6 of them.
44
There are 5,461,512 such combinations.
64
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
It depends on how many digit you are choosing from.
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.
6 of them.
44
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.
How many four digit combinations can be made from the number nine? Example, 1+1+2+5=9.
the answer is = first 2-digit number by using 48= 28,82 and in 3 digit is=282,228,822,822
There are 126 different 5 digit combinations. Note that the combination 12345 is the same as the combination 45312.
There are twelve possible solutions using the rule you stated.