8C3 = 56 of them
28, or 56 counting reversals
There are 125970 combinations and I am not stupid enough to try and list them!
There are 8*7/(2*1) = 28 combinations.
Assuming that the eight numbers are all different and that none of them are zero, the number of combinations is 8C3 = 8!/(3!*5!) where n! = 1*2*3*...*n the number is 8*7*6 / (3*2*1) = 56
8C3 = 56 of them
56 combinations. :)
28, or 56 counting reversals
There are 125970 combinations and I am not stupid enough to try and list them!
There are 8*7/(2*1) = 28 combinations.
Assuming that the eight numbers are all different and that none of them are zero, the number of combinations is 8C3 = 8!/(3!*5!) where n! = 1*2*3*...*n the number is 8*7*6 / (3*2*1) = 56
8C4 = 70
two
There is just 1 combination of 8 numbers taken 8 at a time.
64
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.
nCr = n!/((n-r)!r!) → 49C8 = 49!/((49-8)!8!) = 49!/(41!8!) = 450,978,066 combinations.