That depends on a few factors:
A) How many unique digits can go into each slot?
B) Can a number be repeated? (eg. is 112 invalid because the 1 is used twice?)
C) Does order matter? (eg. is 123 the same as 132)
Assuming n=10 digits and r=3 slots
If repeats are allowed, then we get 10 possible items in the first slot. Since repeats are allowed, we also have 10 in our second slot, and 10 in our third. We don't have to cancel any out because of their ordering.
10 * 10 * 10 = 1000 = 103 = nr
If repeats are not allowed (let's say we have a bag of 10 pieces, each with a number on it. Once you've used a piece, you can't use it again), we can put any of 10 digits in the first slot, any of the 9 remaining in the second slot, and any of the 8 remaining in the third slot:
10 * 9 * 8 = 720 = 10!/7! = n! / ( n - r )!
If order doesn't matter, you'll need to cancel out any combination that is already represented, and there are formulas for that, but explaining it here might be a bit beyond the scope of the question. You'll find a more detailed lesson in combinations and permutations here:
See link below for more lessons on combinations and permutations
10.
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.
You Can Create 999 Number combinations
6 of them.
There are only four combinations but there are 8 permutations.
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
4*3*2*1 = 24 different combinations.
In most 3-number locks, each number ring offers a choice of 10 digits, from 0 to 9. I that case, there are 103 = 1000 combinations.
The number of combinations is 10C4 = 10*9*8*7/(4*3*2*1) = 210
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.
There are 7C4 = 7!/(4!*3!) = 7*6*5/(3*2*1) = 35 combinations.
just intrested in the number combinations * * * * * Number of combinations = 56C6 = 56*55*54*53*52*51/(6*5*4*3*2*1) = 32,468,436
13579
As there are 26 letters in the alphabet. You can calculate the number of combinations by multiplying 26x26x26, giving you the answer 17576.
Just one. * * * * * Depends on how many numbers are on each ring. If there are x numbers, then the total number of combinations (actually they are permutations) is x*x*x or x3.
The number of combinations is nCr = (59*58*57*56*55)/(5*4*3*2*1) = 5006386
Since a number can have infinitely many digits, there are infinitely many possible combinations.