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Unformatted text preview: stage 2: give each girl 2 jellybeans. Then distribute the remaining 252 Â· 4 = 17 jellybeans with no restriction. Using the trick of Stars and Bars , we have ( 10+71 71 ) = ( 16 6 ) choices for stage 1, and ( 17+71 71 ) = ( 23 6 ) choices for stage 2. Then the total number of ways is Â± 16 6 Â²Â± 23 6 Â² = 808 , 383 , 576 . 5. To count the choices with two defective bulbs, we need two stages: stage 1: choose 2 defective bulbs. Number of choices: ( 5 2 ) stage 2: choose 4 eï¬€ective bulbs. Number of choices: ( 15 4 ) . The number of choices with no restriction is ( 20 6 ) . Then our probability is Â± 5 2 Â²Â± 15 4 Â² Â± 20 6 Â² = 455 1292 . 2...
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This note was uploaded on 12/03/2009 for the course CSE 21 taught by Professor Graham during the Spring '07 term at UCSD.
 Spring '07
 Graham
 Algorithms

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