Probability Please help!?!?
posted by Billy .
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.)

The number of different combinations is
C(40,6) or 40!/(6!34!) = 3 838 380 
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) 
if,89,88,45,9,33, were selected from 190, what are the next five numbers to be selected.
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