Question 159217
1a) nCr(40,7)=n!/(n-r)!r!=18,643,560 selections can be made.
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b) nCr(34,7)=5,379,616 selections that wont find a defective one.
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c) 
nCr(6,2)*nCr(34,5)
=15*278256
=4,173,840 will contain 2 defective ones.
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2)containing 1 r: nPr(4,3)=n!/(n-r)!=24 without the permutation of "bae" (not allowed) 3!=6
24-6=18
containing 2r's: 3 ways to arrange 3 letters among the 2 r's: 3*3=9
9+18=27 ans.
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3)We actually have the arrangement of "tatistic" which has 3 t's and 2 i's.
P=n!/r[t]!*r[i]!=8!/3!2!=8*7*6*5*4/2=3360
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Ed