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".
There are 29 ways.
I believe the factors/digits would have to be either 1, 1, 2, 3 (12 combinations) or 1, 1, 1, 6 (4 combinations).
There are 21 combinations.
If the 6 digits can be repeated, there are 1296 different combinations. If you cannot repeat digits in the combination there are 360 different combinations. * * * * * No. That is the number of PERMUTATIONS, not COMBINATIONS. If you have 6 different digits, you can make only 15 4-digit combinations from them.
12 Combinations.
30
6 or 12
about 12
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".
There are 24C12 = 24!/[12!*12!] = 2,704,156 combinations.
252 combinations, :)
There are 29 ways.
I believe the factors/digits would have to be either 1, 1, 2, 3 (12 combinations) or 1, 1, 1, 6 (4 combinations).
To calculate the number of combinations of 6 numbers chosen from 1-36, we can use the combination formula, which is nCr = n! / (r!(n-r)!). In this case, n = 36 (total numbers) and r = 6 (numbers chosen). Plugging these values into the formula, we get 36! / (6!(36-6)!) = 1,947,792 combinations.
There are 1,120,529,256 combinations.
There are: 9C6 = 84 combinations