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
To calculate the combinations for the numbers 2, 6, and 8, we need to use the formula for combinations, which is nCr = n! / r!(n-r)!. In this case, we have 3 numbers (n=3) and we want to choose 2 of them (r=2). So, the combinations would be 3C2 = 3! / 2!(3-2)! = 3. Therefore, the combinations for 2, 6, and 8 are (2, 6), (2, 8), and (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
Well, honey, with 3 numbers to choose from (3, 6, 9), you can make 6 combinations because you can arrange them in any order. So, it's like a little math party with 6 different ways to mix and match those digits. Hope that clears things up for ya!
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.
10C6 = 10*9*8*7/(4*3*2*1) = 210 combinations.