[ March 8, 2007 ]
Kanisher Caldwell's (email) combinatorial problem.
In how many ways can you arrange the numbers 156 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
42 numbers (n) and having all six (m) numbers correct is as follows
or 42! / (36!*6!) =
(1,4050_*10^51) / ((3,7199_*10^41) * (720)) =
So you have 1 chance in 5245786 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 nine
which have to correspond as well.
50!
(505)! . 5!

* 
9!
(92)! . 2!

The left part of the equation amounts to 2118760
and the right side equals 36, so in total you have
1 chance in 76275360 to win the euromillions.
This is 14,5 times less than the regular lottery...
now I know where to participate and try my luck.