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Unformatted text preview: Number of drawers = 20 (corresponding to the donut types) Number of dividers = 19 Number of donuts chosen = 12 Total number of solutions = C 12 + 20 ! 1,12 ( ) = 31 12 " # $ % & = 31 19 " # $ % & = 141120525 (iv) at least two varieties = total possible exactly 1 variety 31 12 ! " # $ % & 20 = 141120505 McCombs Math 381 Boxes 2 (v) at least 6 glazed donuts Number of drawers = 20 (corresponding to the donut types) Number of dividers = 19 Number of donuts chosen = 12 6 = 6 Total number of solutions = C 6 + 20 ! 1,6 ( ) = 25 6 " # $ % & = 25 19 " # $ % & = 177100 (vi) no more than 6 glazed donuts = total at least 6 glazed donuts exactly 6 glazed donuts = 31 12 ! " # $ % & 25 6 ! " # $ % & + 24 6 ! " # $ % & = 141078021...
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This note was uploaded on 06/16/2011 for the course MATH 381 taught by Professor Na during the Summer '11 term at University of North Carolina School of the Arts.
 Summer '11
 NA
 Math

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