algorithm - formula for lotto combinations with fixed numbers? -


usually lottery combinations formula "n! / (k!*(n-k)!)", e.g., 6/49 game "49!/(6!*(49-6)!)"

is there formula calculate same m fixed values (e.g., numbers 1, 2, 3, , 4 fixed) - 2 numbers free choice

i thought formula "(n-m)! / (k!*(n-k-m)!)" doesn't seem ... because m=4 formular wrong (for 6/10) , m=k should 1)

what have posted formula nothing combination.

it says k-combination set of n elements given nck or n! / (k!*(n-k)!).

next, number of combinations of n different things taken m @ time, when k particular objects occur n-mck-m or (n-m)! / (k-m)!*(n-k)!).


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