When the order does not matter, the number ways of choosing the lottery can be calculated using combinations, or C(n,r), read as "n chose r". C(n,r) is defined as n!/(r!(n-r)!), similar to the evaluation problem of your other post.
So in the case of the lottery problem, n=40, r=6, and order does not matter. So "40 choose 6" gives the answer.
Evaluate "40 choose 6" the same way as the other problem:
40!/(6!(40-6)!)=....
Read the following link for more detailed information:
http://en.wikipedia.org/wiki/Combination
A lottery is set up in which players pick six numbers from the set 1, 2, 3, ... , 39, 40. How many different ways are there to play this lottery? (In this game the order in which the numbers are picked does NOT matter.)
1 answer