Solutions from a student (dragged) 12

Solutions from a student (dragged) 12 - Problem 27...

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Unformatted text preview: Problem 27 (counting committees) This is given by the multinomial coefficient 12 3,4,5 = 27720 Problem 28 (divisions of teachers) If we decide to send n1 teachers to school one and n2 teachers to school two, etc. then the total number of unique assignments of (n1 , n2 , n3 , n4 ) number of teachers to the four schools is given by 8 . n1 , n2 , n3 , n4 Since we want the total number of divisions, we must sum this result for all possible combinations of ni , or 8 n1 , n2 , n3 , n4 = (1 + 1 + 1 + 1)8 = 65536 , n1 +n2 +n3 +n4 =8 possible divisions. If each school must receive two in each school, then we are looking for 8 2,2,2,2 orderings. = 8! = 2520 , (2!)4 Problem 29 (dividing weight lifters) We have 10! possible permutations of all weight lifters but the permutations of individual countries (contained within this number) are irrelevant. Thus we can have 10! = 3! · 4! · 2! · 1! 10 3,4,2,1 = 12600 , possible divisions. If the united states has one competitor in the top three and two in 3 the bottom three. We have possible positions for the US member in the first three 1 3 positions and possible positions for the two US members in the bottom three positions, 2 giving a total of 3 3 = 3· 3 = 9, 2 1 ...
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This note was uploaded on 02/25/2011 for the course STAT 418 taught by Professor G.jogeshbabu during the Winter '08 term at Pennsylvania State University, University Park.

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