You can select 9 numbers for the first digit, 8 numbers for the second digit, and 7numbers for the third digit; so 504 (e.g. 9*8*7) different three digit numbers can be written using the digits 1 through 9.
9
10
To calculate the number of 9-digit combinations that can be made from the numbers 1-9 without repetition, we use the permutation formula. Since there are 9 numbers and we are choosing 9 without repetition, the formula is 9P9 = 9! / (9-9)! = 9! / 0! = 9! = 362,880. Therefore, there are 362,880 9-digit combinations that can be made from the numbers 1-9.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
It is: 9C7 = 36
999-111=888 888 3 digit numbers can be made with numbers between 1 - 9
If you mean how many different sets of 7 can be made from 9 numbers, the answer is (9 x 8)/2 ie 36.
10
it is 9 because other numbers are made up by the same numbers already used 1-9 for example
1 - 1 2 - 4 3 - 9 4 - 16 5 - 25 6 - 36 7 - 49 8 - 64 9 - 81
You can select 9 numbers for the first digit, 8 numbers for the second digit, and 7numbers for the third digit; so 504 (e.g. 9*8*7) different three digit numbers can be written using the digits 1 through 9.
3,5,7,8
900000
9
10
that would be 9 to the 7th power i believe.