Posted by **Billy ** on Wednesday, January 12, 2011 at 2:16pm.

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.)

- Probability Please help!?!? -
**Reiny**, Wednesday, January 12, 2011 at 3:52pm
The number of different combinations is

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

- Probability Please help!?!? -
**helper**, Wednesday, January 12, 2011 at 4:11pm
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)

- Probability Please help!?!? -
**Hafiz Baba**, Wednesday, July 8, 2015 at 1:18am
if,89,88,45,9,33, were selected from 1-90, what are the next five numbers to be selected.

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