usually for lottery combinations the formula is "`n! / (k!*(n-k)!)`

", e.g., for a 6/49 game it is "`49!/(6!*(49-6)!)`

"

is there a formula to calculate the same with `m`

FIXED values (e.g., the numbers 1, 2, 3, and 4 are fixed) - 2 numbers are free for choice

i thought the formula is "`(n-m)! / (k!*(n-k-m)!)`

" but it doesn't seem so ... because for `m=4`

the formular is definitely wrong (for 6/10) and for `m=k`

it should be 1)

Answer:

What you have posted as a formula is nothing but Combination.

It says that k-combination from a set of n elements is given by **^{n}C_{k}** OR

`n! / (k!*(n-k)!)`

.Next, the number of combinations of n different things taken m at a time, when k particular objects occur is **^{n-m}C_{k-m}** OR

`(n-m)! / (k-m)!*(n-k)!)`

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