a) Combination of 20 things 8 at a time. 20!/(8!12!) (n things k at a time: n!/(k!(n-k)!) )
b) 0 defects means picking 8 from nondefect pile only: Combination of 17 things 8 at a time. 17!/(8!9!)
c) 3 defects means picking all three defective parts leaving 17 nondefective parts to be selected 5 at a time:
17!/(5!12!)
d) At most one defect means one or zero (the answer to (b)) So let's figure out how many 1 defect selections we can make:
we can make 3 choices for the 1 defect and 7 choices from the 17 nondefect pile:
3 x 17!/(7!10!) (now add part b result)
Note that 2 defects will result from 3 things 2 at a time (3 possibilities) and 17 things 6 at a time:
3!/(2!1!) x 17!/(6!11!)
If you add all the possibilities, they add up to (a)
Hope that helps.
