(8 x 7 x 6 x 5 x 4 x 3)/(6 x 5 x 4 x 3 x 2) = 28 combinations
There are 1,120,529,256 combinations.
There are: 9C6 = 84 combinations
(9 x 8 x 7 x 6 x 5 x 4)/(6 x 5 x 4 x 3 x 2) = 84 combinations.
The number of 4 different book combinations you can choose from 6 books is;6C4 =6!/[4!(6-4)!] =15 combinations of 4 different books.
The answer is 10!/[6!*(10-6)!] where n! represents 1*2*3*...*n Number of combinations = 10*9*8*7*6*5*4*3*2*1/(6*5*4*3*2*1*4*3*2*1) = 10*9*8*7/(4*3*2*1) = 210
There are 8 possible combinations:{},{2},{6},{8},{2,6},{2,8},{6,8} and{2,6,8}There are 8 possible combinations:{},{2},{6},{8},{2,6},{2,8},{6,8} and{2,6,8}There are 8 possible combinations:{},{2},{6},{8},{2,6},{2,8},{6,8} and{2,6,8}There are 8 possible combinations:{},{2},{6},{8},{2,6},{2,8},{6,8} and{2,6,8}
There are 8!/[6!(8-6)!] = 8*7/2 = 28 - too many to list.
5*6*7*8=1680
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
Since there are 3 numbers to choose from, there will be 2^3 = 8 combinations. This includes the null combinations which you may not wish to include in the count.
8C6 = 28
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".
6 for 3-digits, 6 for 2-digits, 3 for 1-digits, and 15 for all of the combinations
There are: 8C6 = 8!/6!(8-6)! = (8 x 7)/(2 x 1) = 28.
27
There are 8*7/(2*1) = 28 combinations.
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.