There are 125970 combinations and I am not stupid enough to try and list them!
There are 8*7/(2*1) = 28 combinations.
8C3 = 56 of them
two
nCr = n!/((n-r)!r!) → 49C8 = 49!/((49-8)!8!) = 49!/(41!8!) = 450,978,066 combinations.
There are 125970 combinations and I am not stupid enough to try and list them!
There are 8*7/(2*1) = 28 combinations.
8C4 = 70
8C3 = 56 of them
two
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.
56 combinations. :)
There are 8!/[6!(8-6)!] = 8*7/2 = 28 - too many to list.
41
Assuming you meant how many combinations can be formed by picking 8 numbers from 56 numbers, we have:(56 * 55 * 54 * 53 * 52 * 51 * 50 * 49)/8! = (7 * 11 * 3 * 53 * 13 * 51 * 25 * 7) = 1420494075 combinations. (Also equal to 57274321104000/40320)