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5 Pages

### C__work_280sol6

Course: CS 280, Spring 2007
School: Cornell
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Word Count: 1953

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to Solutions Homework 6 April 4, 2007 4.3: 35 (b) 40 students play eld hockey, and 1 of those swims, so 39 play eld hockey but dont swim. (d) This question is asking how many students do not play eld hockey. Since there are 1400 students and 40 play eld hockey, 1360 do not play eld 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...

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to Solutions Homework 6 April 4, 2007 4.3: 35 (b) 40 students play eld hockey, and 1 of those swims, so 39 play eld hockey but dont swim. (d) This question is asking how many students do not play eld hockey. Since there are 1400 students and 40 play eld hockey, 1360 do not play eld 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 n k+1 k (n+1)n! = (k+1)k!(nk)! (n+1)! = (k+1)!(nk)! = = (n+1)! (k+1)![(n+1)(k+1)]! (n+1)! (k+1)!(nk)! 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 = (n+1) n . k+1 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: rst 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 ways of doing that. Thus, there are (n + 1) n ways of doing k k that. Alternatively, rst choose the k + 1 objects that will go into boxes A and B ( and then choose the one out of k + 1 that will go into box A. That gives you (k + 1) 1 n1 k+1 n+1 k+1 ) . 4.5: 4 First, heres the argument using algebra. Ill start with the right-hand side: = = = C(n 2, k 2) + 2C(n 2, k 1) + C(n 2, k) (n2)! (n2)! (n2)! + 2 (k1)!(nk1)! + k!(nk2)! (k2)!(nk)! k(k1)(n2)!+2k(nk)(n2)!+(nk)(nk1)(n2)! k!(nk)! (k2 k+2kn2k2 +n2 2kn+k2 n+k)(n2)! k!(nk)! (n2 n)(n2)! k!(nk)! n(n1)(n2)! k!(nk)! = = = 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 denitely 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). Finally, the number of ways of doing (d) is C(n 2, k), since you have to chooose all k out of the n 2 remaining ones. By the Sum Rule, it follows that C(n, k) = C(n 2, k 2) + 2C(n 2, k 1) + C(n 2, k). Note: The homework asked you to do this both ways, even though the book didnt. You only got half marks if you did it one way. You could have told a dierent combinatorial story than the one given here, although the essential features had to be more or less the same. 4.5: 6 First, the algebraic argument: P (n 1, k) + kP (n 1, k 1) (n1)! (n1)! = (n1k)! + k (nk)! = (nk)(n1)!+k(n1)! (nk)! = n(n1)! (nk)! n! = (nk)! = P (n, k) Now the combinatorial argument: We want to arrange k things out of n (the order matters). Suppose the n things are {1, . . . , n} again. Either 1 is in your arrangement or its not. If its not, then you have to arrange k things out of n 1, and there are P (n 1, k) ways of doing that. If it is, you now have to think of where 1 goes in the arrangement (remember, order matters). It could be the rst element, the second, . . . , 2 or the kth element. Bottom line: you have k choices of the position of 1 in the list. Once youve decided that, you have to arrange the other k 1 things out of n1; this gives you kP (n 1, k 1). By the Sum Rule, we get that P (n, k) = P (n 1, k) + kP (n 1, k 1). 4.6: 2(a) 4 (1.01)4 = k=0 4 4k 1 (.01)k k = 1 + 4(.01) + 6(.01)2 + 4(.01)3 + (.01)4 = 1 + .04 + .0006 + decimals with at least 4 leading 0s 1.0406 n nk k , but excepting the rst two terms of this summation, 4.6: 6 (a + )n = n k=0 k a 2 all terms contain or higher powers of , and thus have a negligible eect on the sum. Thus (a + )n an + n an1 . In Problem 5, this yields the approximation (1 + .01)12 112 + 12(.01)(111 ) = 1.12 Comment Its surprising that many students didnt answer How this problem is relevant to the previous problem. You lost 1 point if you didnt. Problem 5 is a specic case of this problem, and you are expected to apply your formula to it. Most students, on the other hand, pointed out the correspondence of the two problems. There is some ambiguity in the question so I accepted any reasonable answer. 4.6: 8 n (1 + x)n = k=0 n n nk k 1x k nk x k n2 n3 x+ x + + xn 2 3 = k=0 = 1 + nx + 1 + nx since all deleted terms are nonnegative for x 0. If n = 0, there is only one term in the expansion of (1 + x)n , but the claim, (1 + x)n 1 + nx, is still correct because it says 1 1. Comment Some students argued that all the deleted terms were positive. This is not necessarily true (for example, x = 0). You lost 1 point if you proved (1 + x)n > 1 + nx instead (1 of + x)n 1 + nx. 3 4.7: 4 At most 43 . Note that it is backwards to think of the positions in the string as urns, and the bases as balls, because each position must be assigned exactly one base, not the other way around. 4.7: 8 u b . b urns must be chosen to hold a ball. 4.7: 9 There are D+d ways of getting a sequence of D + d dots and dashes, of which d there are d dashes and D dots: you simply choose the d places out of D + d where the dashes will go. Now heres a way of getting the answer that involve balls and urns: Consider the number of ways of putting D indistinguishable balls into d + 1 distinguishable urns. According to what we discussed in class, this can be done in D+d ways. Theres a 1-1 D correspondence between each choice and a way of arranging the dots and dashes: The number of balls in urn k is the number of dots between the (k 1)st dash and the kth dash. Thus, the number of balls in the rst urn is the number of dots at the beginning of the string, before you see a dash, and the number of balls in the last urn (the (d + 1)st) is the number of dots after the last dash (the dth dash). It should be pretty clear that this gives a 1-1 correspondence: For every way of putting D balls into d + 1 urns, theres a unique sequence of dots and dashes, and for every sequence of dots and dashes, theres a unique way of putting balls into d + 1 urns. 4.7: 15(b) You can think of the squares on a face as indistinguishable balls and the colors as urns. There are thus 9 indistinguishable balls and 6 distinguishable urns (since there are six faces on the cube and a dierent color for each face). Thus, a color frequency of 5 red squares, 3 blue squares, and 1 yellow square corresponds to putting 5 balls in the red urn, 3 in the blue urn, and 1 in the yellow urn. By the formula proved in class, there are 14 possible color frequencies. 9 4.10, 2 The answer should not change. We still have 5 holes (colors), and 6 pigeons (socks) which are needed to guarantee getting a matching pair of socks. 4.10, 8 NOTE: YOU SHOULD READ OVER THIS SOLUTION CAREFULLY EVEN IF YOU RECEIVED FULL CREDIT FOR THE PROBLEM! This was a dicult problem; few people received full credit. First o, heres a false proof of the result given in the problem. It is sometimes instructive to see why certain methods are not valid ways of reasoning (remember faulty induction?): Proof: Partition the set {1, 2, 3, ..., 2n} into subsets of the form {k, 2k} for k = 1, 2, . . . , n. Certainly, the worst case would correspond to picking as many numbers as possible from dierent subsets, because then there is no guarantee that the elements we 4 pick are multiples of one another. But we are choosing n + 1 distinct elements from n sets, meaning that there are two elements belonging to the same subset, hence one of the n+1 elements is twice the other. Why this proof is incorrect: The fallacy in the above proof is that the subsets of the form {k, 2k} do not cover the set {1, 2, . . . , 2n}. That is, they are not disjoint subsets. Notice that for n = 4, we have {1, 2}; {2, 4}; {3, 6}; {4, 8}}. The elements 2 and 4 pop up twice and elements 5 and 7 do not appear at all. Therefore, this proof is buggy. Bigtime. Correct proof: The set {1, 2, . . . 2n} contains n odd integers: 1, 3, . . . (2n 1). For each of the n + 1 elements that you choose, write each in the form 2x y where x is the highest power of 2 that divides the element and y is the largest odd factor of the element (note that if the element a is odd, then x = 0 and y = a). There are clearly only n possibilities for y (since the set contains n odd integers), but we are picking n + 1 elements, which means (by the pigeonhole principle) that at least two of these elements must have y-values which coincide. Now assume that these elements with the same yvalues are a and b. Then we assume, WLOG1 that a > b. Then a = 2x y and b = 2x y for some integers x and x with x > x . Then clearly a = 2xx b, and we see that a must be an integer multiple of b. Q.E.D.2 Kleinberg problem: For every xed ordering of requests from the rst and second program, there are C(n + m, n) (or, equivalently, C(m + n, m)) possible interleavings of the programs. To see this, note that there will be m + n requests in a combined stream. Think of there as being m + n slots where a request can go. An interleaving amounts to choosing the m slots where the request issued by the rst program will go. (For example, if m = 3 and n = 2, the rst program issues requests r1 ; r2 ; r3 , and the second program issues requests s1 ; s2 , then one interleaving is r1 ; r2 ; s1 ; r3 ; s2 . This means that the rst, second, and fourth requests are from the rst program; that is, weve chosen slots 1, 2, 4 for the rst program. Similarly, for the interleaving r1 ; s1 ; s2 ; r2 ; r3 , then the rst program gets slots 1, 4, and 5. There are C(m + n, m) ways of choosing m slots out of m + n. This is a standard notation in proofs. WLOG means Without loss of generality. Again, this is another typical notation used: QED literally means quod erat demonstratum, which is Latin for that which was to be proved. I personally would NEVER write this at the end of a proof if you are not absolutely certain that the proof is correct. Some people tried doing this for proofs that were obviously incorrect. No, that does not win you more points. 2 1 5
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