For the permutation the equation is (8!)/(5!) assuming there is no replacement. Which is 8*7*6 = 336 For combinations the equation is (8!)/((5!)(8-5)!) = (8!)/((5!)(3)!) = (8*7*6)/(6) = 56
There are 8*7/(2*1) = 28 combinations.
it depends on which 8 numbers your talking about
With 2 6-sided dices there are 6x6=36 combinations (6 for the first dice, 6 for the second dice) Let put the question this other way: how many combinations with a total more than 8, namely 9, 10, 11 or 12 ? 9: 4 combinations: 4-5 or 5-4 or 6-3 or 3-6 10: 3 combination: 5-5 or 6-4 or 4-6 11: 2 combinations: 6-5 or 5-6 12: 1 combination: 6-6 So there are 10 combinations for "more than 9" thus there are 36-10=26 combinations for "less or equal to 8".
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
For the permutation the equation is (8!)/(5!) assuming there is no replacement. Which is 8*7*6 = 336 For combinations the equation is (8!)/((5!)(8-5)!) = (8!)/((5!)(3)!) = (8*7*6)/(6) = 56
3*5*8 = 120
5*6*7*8=1680
41
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
(8 x 7 x 6 x 5 x 4 x 3)/(6 x 5 x 4 x 3 x 2) = 28 combinations
There are 8*7/(2*1) = 28 combinations.
7,893,600 (seven million, 8 hundred ninety-three thousand, 600) combinations in English.
it depends on which 8 numbers your talking about
fifty * * * * * What sort of rubbish answer is that? There are 10C5 = 10*9*8*7*6/(5*4*3*2*1) = 252 combinations.
With 2 6-sided dices there are 6x6=36 combinations (6 for the first dice, 6 for the second dice) Let put the question this other way: how many combinations with a total more than 8, namely 9, 10, 11 or 12 ? 9: 4 combinations: 4-5 or 5-4 or 6-3 or 3-6 10: 3 combination: 5-5 or 6-4 or 4-6 11: 2 combinations: 6-5 or 5-6 12: 1 combination: 6-6 So there are 10 combinations for "more than 9" thus there are 36-10=26 combinations for "less or equal to 8".
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.