algorithm,math,combinations , formula for lotto combinations with fixed numbers?


formula for lotto combinations with fixed numbers?

Question:

Tag: algorithm,math,combinations

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 nCk 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-mCk-m OR (n-m)! / (k-m)!*(n-k)!).


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