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C__work_280sol6

C__work_280sol6 - Solutions to Homework 6 April 4 2007 4.3...

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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 don’t 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 1000 /n 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 inclusion-exclusion principle, this is: 1000 / 3 + 1000 / 5 + 1000 / 7 - 1000 / 15 - 1000 / 21 - 1000 / 35 + 1000 / 105 = 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 it’s worth, here’s a combinatorial argument (although it wasn’t 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
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4.5: 4 First, here’s the argument using algebra. I’ll start with the right-hand 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 k - 2 things out of the remaining n - 2. The number of ways of doing (b) – choosing k things that include 1 and not 2 – is C ( n - 2 , k - 1) (1 is definitely in, 2 is out, and you need k - 1 more things out of the n - 2 remaining). Similarly, the number of ways of doing (c) is C ( n - 2 , k - 1).
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