The general formula is to square the number of digits it could be (in this case 5) by how many spaces there are. So in this case it would be 53 = 125 possible combinations.
120 5-digit numbers can be made with the numbers 12345.
1,956 different numbers can be made from 6 digits. You can calculate this by using the permutation function in a summation function, like this: Σ6k=1 6Pk = 6P1+6P2+...+6P5+6P6 What this does is calculate how many 1 digit numbers you can make from 6 digits, then how many 2 digit numbers can be made from 6 digits and adds the amounts together, then calculates how many 3 digit numbers can be made and adds that on as well etc.
0 you can only have 3 layers since you have only 3 numbers so without repeating you would only have 3 layers
a lot
The answer is 10C4 = 10!/[4!*6!] = 210
2
210
9
There are 4 options for the hundreds digit (3, 4, 5, or 6), and 4 options for the tens digit (including the possibility of repeating the hundreds digit). Similarly, there are 4 options for the units digit. Therefore, the total number of 3-digit numbers that can be formed using the digits 3, 4, 5, and 6 with repetition allowed is 4 x 4 x 4 = 64.
You can make 4*3*2/2 = 12 numbers.
5040 numbers can be made.
-- If the same digit may be repeated, then 64 can be made. -- If the same digit may not be repeated, then 24 can be made.