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Unformatted text preview: Solutions to Homework 6 April 4, 2007 4.3: 35 (b) 40 students play field hockey, and 1 of those swims, so 39 play field hockey but dont swim. (d) This question is asking how many students do not play field hockey. Since there are 1400 students and 40 play field hockey, 1360 do not play field hockey. 4.3: 36 First, there are b 1000 /n c positive integers less than 1000 which are multiples of n (see if you can convince yourself of this). We must count all the multiples of 3, 5, or 7, but must not double count the numbers that are multiples of 3 AND 5 or 3 AND 7, etc. By the inclusionexclusion principle, this is: b 1000 / 3 c + b 1000 / 5 c + b 1000 / 7 c  b 1000 / 15 c  b 1000 / 21 c  b 1000 / 35 c + b 1000 / 105 c = 333 + 200 + 142 66 47 28 + 9 = 543 . 4.5: 2 Algebra: We rewrite both sides without changing their values until we get the same expansion. n +1 k +1 n +1 k +1 n k = ( n +1)! ( k +1)![( n +1) ( k +1)]! = ( n +1) n ! ( k +1) k !( n k )! = ( n +1)! ( k +1)!( n k )! = ( n +1)! ( k +1)!( n k )! For what its worth, heres a combinatorial argument (although it wasnt necessary to do this to get full credit): Identities involving division are hard to explain by stories having integer answers, so rewrite the equation to be proved as ( k +1) n +1 k +1 = ( n +1) n k . Now suppose you have n + 1 objects and two boxes, say A and B . You want to put one object in box A and k objects in box B . There are two ways of doing that: first choose the object for box A ( n + 1 choices) and choose the k objects for box B out of the remaining n ; there are n k ways of doing that. Thus, there are ( n + 1) n k ways of doing that. Alternatively, first choose the k +1 objects that will go into boxes A and B ( n 1 k +1 ) and then choose the one out of k +1 that will go into box A . That gives you ( k +1) n +1 k +1 . 1 4.5: 4 First, heres the argument using algebra. Ill start with the righthand side: C ( n 2 , k 2) + 2 C ( n 2 , k 1) + C ( n 2 , k ) = ( n 2)! ( k 2)!( n k )! + 2 ( n 2)! ( k 1)!( n k 1)! + ( n 2)! k !( n k 2)! = k ( k 1)( n 2)!+2 k ( n k )( n 2)!+( n k )( n k 1)( n 2)! k !( n k )! = ( k 2 k +2 kn 2 k 2 + n 2 2 kn + k 2 n + k )( n 2)! k !( n k )! = ( n 2 n )( n 2)! k !( n k )! = n ( n 1)( n 2)! k !( n k )! = C ( n, k ) Now, the combinatorial argument: C ( n, k ) is the number of ways of choosing k things from n . Suppose the n things are { 1 , . . . , n } . Of the k things you choose, either (a) you choose both 1 and 2, (b) you choose 1 and not 2, (c) you choose 2 and not 1, or (d) you choose neither 1 nor 2. The number of ways of doing (a) choosing k things that include both 1 and 2 is C ( n 2 , k 2) (you have to choose...
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This note was uploaded on 09/04/2008 for the course CS 280 taught by Professor  during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 

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