0000, 0009, 0090, 0099, 0900, 0909, 0990, 0999, 9000, 9009, 9090, 9099, 9900, 9909, 9990, 9999
im not geeky enough to answer that
90
7
The number of possible 3 digit combinations you can make out of 1-9 with outrepeated digits is:9C3 = 9!/(3!(9-3)!) = 84
Not sure what a "didget" is. It is possible to make 18 5-digit numbers.
You can make 24 4-digit numbers. 1111 2222 3333 4444 5555 6666 1234 2345 3456 1356 6543 5432 4321 1432 ok i give up!
There are 5*4*3*2 = 120 such numbers.
There are many combinations possible The most popular combination is 2,3,5,17,23.
4578, 4758, 5478, 5748, 5874, 5784, 7584, 7854, 7458, 7548, 8754, 8574
Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.
106 or a million.
Just one. In a combination, the order of the digits does not matter.
There should be 125 combinations, but if you can only use the numbers once for each 3 digit number, then it should be 120. It's 125 because if the digits could be used more than once for each number then its the number on possible numbers (so there's 5 in this case, 2, 3, 5, 8, and 9) to the power of how big the digit is (in this case it's 3) It's 120 because if the digits could only be used once in each number then the combinations would be the number of possible numbers (again, 5) factorial, so 5!