WRONG ANSWER NB !!If repeated numbers are permitted the answer is 8x8x8x8x8=32768 (85)
If the numbers can only be used once in each combination, the answer is 8x7x6x5x4= 6720, where ABCDE has [n!/ (n-r)!] or [8! / 3!] combinations for 8 numbers used 5 at a time.
NONONO
NO THIS IS NOT THE RIGHT FORMULA IT SHD BE [n ! / (n-r)! x r! ] = ONLY 56 COMBINATIONS where each number can only be used once.
8x7x6x5x4x3x2x1 / (3x2x1) ( 5x4x3x2x1) = 56
from napoleon solo
5
45.
There are 5*4*3*2 = 120 such numbers.
Just one. In a combination, the order of the digits does not matter.
106 or a million.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
90
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.
There are 5*4*3*2 = 120 such numbers.
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.