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

In how many ways can you arrange the numbers 1-56 in packs of six ?

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

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.

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.