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.)

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

To find the number of different ways to play this lottery, we need to consider combinations since the order of the numbers does not matter.

In this case, we need to choose 6 numbers out of 40 available numbers. The formula for finding the number of combinations is given by:

C(n, r) = n! / (r! * (n-r)!)

Where n is the total number of items to choose from and r is the number of items to be chosen.

In our case, n = 40 (the total number of available numbers) and r = 6 (the number of numbers to be chosen).

Therefore, the number of different ways to play this lottery is:

C(40, 6) = 40! / (6! * (40-6)!)

Now, let's calculate this:

40! = 40 * 39 * 38 * 37 * 36 * 35 * ... * 3 * 2 * 1
6! = 6 * 5 * 4 * 3 * 2 * 1
(40-6)! = 34 * 33 * 32 * 31 * ... * 3 * 2 * 1

Simplifying the above expression:

C(40, 6) = (40 * 39 * 38 * 37 * 36 * 35 * ... * 3 * 2 * 1) / ((6 * 5 * 4 * 3 * 2 * 1) * (34 * 33 * 32 * 31 * ... * 3 * 2 * 1))

Calculating this expression will give us the number of different ways to play the lottery.