There are 1285C4 = 1285*1284*1283*1282/(4*3*2*1) = 113076300485 combinations.
If repeats are allowed than an infinite number of combinations is possible.
61
There are only five combinations: 1234, 1235, 1245, 1345 and 2345.
im assuming that any charcter can be a number or a letter: (24letters*10 possible numbers)^(4 digits)= 3317760000 possible combinations.
10,000
Since a number can have infinitely many digits, there are infinitely many possible combinations.
35
There are infinitely many numbers and so infinitely many possible combinations.
2^n possible combinations
If repeats are allowed than an infinite number of combinations is possible.
61
The number of combinations - not to be confused with the number of permutations - is 2*21 = 42.
Just 4: 123, 124, 134 and 234. The order of the numbers does not matter with combinations. If it does, then they are permutations, not combinations.
As there are 26 letters in the alphabet. You can calculate the number of combinations by multiplying 26x26x26, giving you the answer 17576.
There are only five combinations: 1234, 1235, 1245, 1345 and 2345.
If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.
If the order of the numbers are important, then this is a simple combination problem. There are 10 possible numbers to choose from for the first number. Then there are 9 options for the second number. Then there are 8 options for the third, and so on. Thus, the number of possible combinations can be calculated as 10x9x8x7x6x5. This comes out at 151,200 possible combinations.