5
45.
There would be 9*9*9*9 or 6561 combinations.
-8
Exactly 3,628,800, or 10!.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
You would get 4!/2! = 12 combinations.
There are 840 4-digit combinations without repeating any digit in the combinations.
7
56 combinations. :)
Any 6 from 51 = 18,009,460 combinations
5
45.
To calculate the number of 3-digit combinations using numbers 1-6, including 1 and 6, we need to consider the total number of options for each digit. Since we are including 1 and 6, there are 6 options for the hundreds place (1-6), 6 options for the tens place (1-6), and 6 options for the ones place (1-6). Therefore, the total number of 3-digit combinations is 6 x 6 x 6 = 216.
There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.
90