9 times 9 times 9 times 9 times 9 times 9 times 9 times 9 times 9 =
Imagine it's a combination lock with 9 tumblers. Each tumbler has 9 "positions" - the digits from 1 to 9. So you multiply the number of digits [9] against itself [9] for as many tumblers are in the combination (9).
Tumbler # 1 2 3 4 5 6 7 8 9
Number of
possible
digits 9 9 9 9 9 9 9 9 9
1,000. The list looks just like the counting numbers from 000 to 999 .
Any 6 from 51 = 18,009,460 combinations
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.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
If you are able to repeat numbers you would take 6 * 6 * 6 = 216 combinations. If you are not able to repeat numbers you would take 6 * 5 * 4 = 120 combinations.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
Exactly 3,628,800, or 10!.
There are 38760 combinations.
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.
It is: 9C7 = 36
66
15
10,000
10!/3! = 604800 different combinations.
that would be 9 to the 7th power i believe.
There are 1140 five digit combinations between numbers 1 and 20.
Through the magic of perms and coms the answer is 729