The first digit can be any one of the ten symbols, 0 thru 9. For each of those . . .
The second digit can be any one of the 9 that weren't used in the first place. For each of those . . .
The third digit can be any one of the 8 that you haven't used yet. For each of those . . .
The fourth digit can be any one of the 7 that haven't been used yet.
Total number of possible arrangements = (10 x 9 x 8 x 7) = 5,040
1000
66
In other words, how many 4 digit combination locks are there using the digits 0-9 on each wheel. There are 10×10×10×10 = 10⁴ = 10,000 such combinations.
There are 35C4 = 35*34*33*32/(4*3*2*1) = 52,360 combinations.
This question needs clarificatioh. There are 4 one digit number combinations, 16 two digit combinations, ... 4 raised to the n power for n digit combinations.
There is only one combination. In a combination the order of the numbers does not matter so the only combination is 0123456789. This is the same as 1326458097
How many numbers per combination?
There are 360 of them.
Including the null combination, there are 2^14 = 16384 combinations.
There is 1 combination of all ten numbers, 10 combinations of one number and of nine numbers, 45 combinations of two or eight numbers, 120 combinations of three or seven numbers, 210 combinations of four or six numbers and 252 combinations of five numbers. That is 1023 = 210 - 1 in total.
There is only one combination since the order of the numbers in a combination does not matter.
1000
10!/3! = 604800 different combinations.
66
15
10,000
575757