There are 6C3 = 20 such combinations.
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.
There are 7C4 = 7!/(4!*3!) = 7*6*5/(3*2*1) = 35 combinations.
13579
There is 345, 354, 435, 453, 543, 534.
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.
All the numbers from 000 to 999 (inclusive).
There are 6C3 = 20 such combinations.
10.
All the possible digits (10 of them; 0-9) are multiplied by themselves by the number of digits that can be shown in the lock. (3) This is 103, or 1,000. This certainly shows why guessing is not a good way to break into a numerical lock, especially since three is a rather low number of digits for one!
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 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.
You Can Create 999 Number combinations
Assuming that repeated numbers are allowed, the number of possible combinations is given by 40 * 40 * 40 = 64000.If repeated numbers are not allowed, the number of possible combinations is given by 40 * 39 * 38 = 59280.
There are: 12345C3 = 3.134847985*1011
Oh, dude, let me break it down for you. So, you've got 44 numbers to choose from, right? And you're looking for 3-number combinations? Well, you can calculate that using the formula for combinations, which is like 44 choose 3. So, the answer is 44 x 43 x 42 divided by 3 x 2 x 1, which equals... wait, you want me to do the math too? Come on, man, I'm a comedian, not a human calculator!
6 of them.