if you can start with 0, there are 5 x 4 x 3 x 2 = 120 combinations. if you can't start with 0 then 4 x 4 x 3 x 2 = 96 combinations
All the numbers from 0 to 999 so 1000 numbers However if you mean only using each number once the answer is about 720 * * * * * No. The above answer refers to the number of permutations. Permutations are NOT the same as combinations as anyone who has studies any probability theory can tell you. The number of combinations is 9C3 = 9*8*7/3*2*1 = 84
24
0000-9999 (10x10x10x10 or 104) = 10,000 possible combinations allowing for repeated digits. If you are not able to repeat digits then it's 10 x 9 x 8 x 7 or 5,040 possible combinations without repeated digits.
7
There are 45 combinations.
There are a huge number of combinations of 5 numbers when using the numbers 0 through 10. There are 10 to the 5th power combinations of these numbers.
9000
There are only 10 combinations. In each combination one of the 10 digits is left out.
If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.
10,000 combinations.
10P4 = 5040
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.
If every number can be used as many times as you like, there are 104 = 10000 different combinations. If each number can only be used once, there are 9!/(9 - 4)! = 5040 combinations.
Since there are only 3 digits available, repetition must be allowed. In that case, there are 30 combinations.
9999
The first digit can be one of five integers (1 to 5) The second - fourth digits one of six integers (0 to 5). So: number of (valid) combinations is 5*6*6*6=1080 * * * * * This gives the number of permutations, not combinations. I am working on the latter.