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.)

Combination of 40 taken 6 at a time

nCr = nPr/r!
nPr = n!/(n - r)!
nCr = (n!/(n - r)!) /r!
n = 40, r = 6
40C6 = 40P6/6! = 40!/34! / 6!
= 40!/(6!*34!)
= 40*39*38*37*36*35/(6*5*4*3*2*1)
= 2,763,633,600/720
= 3,838,380

I am not a tutor but this is right IF this is a combination of 40 taken 6 at a time (I checked my math in an online calculator)

The number of different combinations is

C(40,6) or 40!/(6!34!) = 3 838 380

if,89,88,45,9,33, were selected from 1-90, what are the next five numbers to be selected.

To find the number of different ways to play this lottery, we need to calculate the number of combinations possible. Since the order of the numbers does not matter, we are dealing with combinations rather than permutations.

To calculate the number of combinations, we can use the combination formula:

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

In this case, we have a set of 40 numbers to choose from, and we need to pick 6 of them. So we can use the combination formula:

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

Now, let's calculate:

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

Plugging these values into the formula, we have:

C(40, 6) = 40! / (6!(40 - 6)!)
= 40! / (6! * 34!)

Now, we can simplify the calculation:

40! = 40 * 39 * 38 * 37 * 36 * 35 * 34!
6! = 6 * 5 * 4 * 3 * 2 * 1

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

Canceling out the 34! terms:

C(40, 6) = (40 * 39 * 38 * 37 * 36 * 35) / (6 * 5 * 4 * 3 * 2 * 1)

Now, we can simplify this further by canceling out common factors:

C(40, 6) = (8 * 13 * 19 * 37 * 36 * 7) / (6 * 4 * 2)

Calculating this expression gives us the final answer:

C(40, 6) = 3,838,380

Therefore, there are 3,838,380 different ways to play this lottery.