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!
20From Rafaelrz,The number of three digit permutations you can get from the four different digits3. 6. 7. 9, is given by:4P3 = 4!/(4-3)! = 4x3x2x1/1 = 24
The first digit can be any one of 1,2,3,4,5. With each choice of digit1, the second digit can be any one of 1,2,3,4,5 so making 5*5 two-digit numbers. With each of these, the third digit can be any one of 1,2,3,4,5 so making 5*5*5 three-digit numbers and so on. The total number is 55 = 3125.
ccsndf
-5674
8 of them.
it would be 321 312 231 213 123 132
There are no three didgit numbers but there are 63 three digit numbers.
4*3*2 = 24 of them.
There are 5*4*3 = 60 such numbers.
8
111, 113, 131, 133, 311, 313, 331, 333
55----------------------------------------------------------------------------------------------From Rafaelrz.You can make, 5! = 120, five digit numbers using 1,2,3,4 and 6.
6, as long as they are all different
I take it that you want to make three digits numbers with 8,7,3, and 6 without repetition. The first digit cane be selected from among 4 digits, the second from 3 digits, the third digit from 2, hence the number of three digit numbers that can be formed without repetition is 4 x 3 x 2 = 24
10999
24
144