Question 159217: 1) A shipment of 40 laptop computers containing 6 defective ones. The shipping department selects seven of these laptops and rejects the entire shipment if one or more are defective.
a) How many selections can be made?
b) How many of these selection will contain no defective computers?
c) How many selection will contain at least two defective ones?
2) By playing Scrabble we ended up with the five remaining letters B, A, R, R, E. In how many ways can three of these five letters be arranged in a row if at least one of the letters is R?
3) In how many ways can the letters of the word STATISTICS be arranged in a similar text where the first and the last letters of the word are fixed but the remaining letters inside are arranged arbitrarily?
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! 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
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