Sisplat said:Firstly, 8*7*2 is not equal to 128.
My calculation is:
Firstly for numbers ending in 0,
9*8 possible ways of choosing 2 digits.
For numbers ending in 5,
8*7, since we want to exclude 0
9*8 + 8*7 = 128.
A 3-digit number permutation problem is a math problem that involves finding all possible unique combinations of three digits to create a number. For example, if the digits are 1, 2, and 3, the possible 3-digit numbers are 123, 132, 213, 231, 312, and 321.
To solve a 3-digit number permutation problem, you can use the following steps:
The formula for calculating the number of permutations in a 3-digit number permutation problem is n! / (n - r)! , where n is the total number of digits and r is the number of digits in each permutation. In this case, n = 3 and r = 3, so the formula becomes 3! / (3 - 3)! = 3! / 0! = 3 x 2 x 1 = 6.
Yes, repetitions can be used in a 3-digit number permutation problem. This means that the same digit can be used more than once in a single permutation. For example, if the digits are 1, 2, and 3, the number 111 would be a valid permutation.
3-digit number permutation problems have various real-world applications, such as: