[ March 8, 2007 ]
A combinatorial problem
Kanisher Caldwell (email)

Example, without the numbers repeating themselves, i.e.

{1,2,3,4,5,6}
{1,6,7,9,33,42}
{...}

If the order is taken into account,
and e.g. {1,6,7,9,33,42} {1,33,42,9,6,7} which
are two different but equally valid groups,
how does that reflect in the final count ?

Maybe this story is already a partial solution to your problem :

The formula to calculate your chances to win the belgian lottery with
45 numbers (n) and having all six (m) numbers correct is as follows

n! (n-m)! . m!

or 45! / (39!*6!) =
So you have 1 chance in 8145060 to win with one grid !

Redoing the exercice for the euromillions lottery is
just a bit more elaborated. The grid contains 50 numbers of which
5 need to be correct and then there are two stars from twelve
which have to correspond as well.

50! (50-5)! . 5!

*

12! (12-2)! . 2!

The left part of the equation amounts to 2118760
and the right side equals 66, so in total you have
1 chance in 139838160 to win the euromillions.

This is 17,168... times less than the regular lottery...
now I know where to participate and try my luck.