Answer: (a) Since we cannot repeat men on the committee, and the order we select them in does not matter, ( )


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1 1. (Chapter 1 supplementary, problem 7): There are 12 men at a dance. (a) In how many ways can eight of them be selected to form a cleanup crew? (b) How many ways are there to pair off eight women at the dance with eight of those 12 men? Answer: (a) Since we cannot repeat men on the committee, and the order we select them in does not matter, ( ) (b) One way to think about this there are 12 choices for who gets the first woman, 11 choices for who gets the second, and so on, down to 5 choices for who gets the eighth. We multiply because we have to match up each woman. Another way to think about it is that we re again selecting 8 men without repetition, but now order matters, because choosing Adam for Alice and Bob for Betty is different than choosing Bob for Alice and Adam for Betty. Therefore we can use the formula P (12, 8) = 12!. 8! Comments: At second glance, it seems like the book wasn t very precise in laying out this problem. In particular, you had to assume that there were only eight women to be paired off. If a problem like this is on the exam and you re not sure what assumptions you should be making, ask. On the exam, you d have gotten full credit for just writing down the answer (it s a very straightforward problem). However, showing work would have made it more likely for you to get partial credit if you made a mistake. 2. (Chapter 1 supplementary, problem 10): Mr. and Mrs. Richardson want to name their new daughter so that hew initials (First, middle, and last) will be in alphabetical order with no repeated initial. How many triples of initials can occur under these circumstances? Answer: The trick here is to think about the Richardsons as first choosing three letters to use as the initials, then sorting the letters in alphabetical order. There are ( ) 26 3 ways to choose the three letters, and only one way to put them in order. 3. (Chapter 1 supplementary, problem 12): In how many ways can a teacher distribute 12 different science books among 16 students if (a) no student gets more than one book? (b) the oldest student gets two books but no other student gets more than one book? Answer: (a): Method 1: We re choosing one student for each science book. The problem states that we re not allowed to repeat ( no student gets more than one book ) and order matters ( different books). This means we use P (n, k), which in this case is P (16, 12). Method 2: Think of the teacher as first choosing 12 students to give books to, then giving out the books. There are ( 16 12) ways of choosing the students, and 12! ways of assigning the books, so ( 16 12) 12!. (b): We first give two books to the oldest student, then give out the remaining ten. There are ( ) 12 2 ways to choose two books for the older student, and (for the same reason as part (a)) P (15, 10) ways to give ten books to the remaining students. The answer is ( ) 12 2 P (15, 10). 4. (Chapter 1 supplementary, problem 16a): Find the coefficient of x 2 yz 2 in the expansion of [(x/2) + y 3z] 5 1
2 Answer: By the binomial theorem, the coefficient is 5! 2!1!2! ( 1 2 )2 ( 3) 2. As usual, it s okay if you just leave your answer like that. 5. (Section 4.1, problem 15): Prove that for every positive integer n > 4, we have that n 2 < 2 n. Answer: We use induction. We re proving something for n > 4, so our base case is the smallest integer satisfying that, n = 5. In that case, 25 = 5 2 < 2 5 = 32. Now for the inductive step. We assume that n 2 < 2 n, and want to prove that (n + 1) 2 < 2 n+1. There s a couple ways to prove this. One is to note that (n + 1) 2 = n 2 + 2n + 1 n 2 + 3n( since 1 < n) n 2 + n 2 ( since 3 < n, so 3n < n 2 ) 2 n + 2 n = 2 n+1 ( using our inductive hypothesis ) 6. Find a pair of integers (x, y) for which 19x + 47y = 200. Answer: We first use the Euclidean Algorithm to compute GCD(19, 47). 47 2(19) = (9) = 1 9 9(1) = 0 so the GCD is 1. Now we reverse to write 1 as a combination of 19 and 47: 1 = 19 2(9) = 19 2(47 2(19)) = 19 2(47) + 4(19) = 5(19) 2(47). Then we multiply the equation 5(19) 2(47) = 1 by 200 to get 1000(19) 400(47) = 200. Therefore (1000, 400) is a solution. 7. (Section 5.5, problem 4): Let S = {3, 7, 11, 15, 19,..., 95, 99, 103}. How many elements must we select from S to insure that there will be at least two whose sum is 110? Answer: Divide S into the pairs {7, 103}, {11, 99},... {51, 59}, {55}, {3}. There are 14 pairs total, so if we select 15 elements than by the Pigeonhole principle we ve selected two elements from the same subset, so we get a sum of 110. Comment: Again, the book was a bit unclear in its statement of the problem. I assumed it to mean that we shouldn t select the same element twice, and that the two whose sum is 110 must be different. Again, if you aren t sure during the exam about something like this, ASK! 8. (based on Section 5.5, problem 13): Alice and Bob are playing a game. Alice gives Bob a list of five positive integers, each of which is at most 9. Bob wins if he can find two different 2
3 subsets of that list having the same sum. For example, if Alice chooses 1, 2, 4, 5, 7, Bob can choose {1, 2, 4} and {7}, both of which add to 7. Show that no matter what set of 5 integers Alice chooses, Bob can still win. Answer: Suppose we could show that there are more subsets for Bob to choose from then there are possible sums in the set of numbers that Alice chose. We d then be done by the pigeonhole principle. So let s see if this works. There are 2 5 = 32 choices for Bob s subsets. The largest sum possible for Bob is = 35, and the smallest possible sum is 0 (if he leaves everything out. Unfortunately, this is too large. So we ll do a bit of a trick instead. Suppose we further restricted Bob to only choose subsets of size between 1 and 3. Now there are ( ( 5 1) + 5 ( 2) + 5 3) = 25 choices of subsets Bob can take. However, now the largest possible sum is = 24 instead of 35, and the smallest possible sum is 1. Since we have 25 subsets with at most 24 different sums, two subsets have the same sum by the Pigeonhole principle. But this restriction only makes things harder for Bob. Comment: This problem was harder than I had thought (An arithmetic mistake I made when solving it led me to initially believe that the trick wasn t necessary), and probably a bit hard for the final exam. A more reasonable problem would have been to have Alice only allowed to choose numbers between 1 and Find the number of ways of arranging the letters in MAT HEMAT ICS so that no pair of consecutive letters appears. Answer: We use inclusionexclusion. The conditions we want to avoid are: c M : the two M s are consecutive c A : the two A s are consecutive c T : the two T s are consecutive. We have N = S 0 = 11!. For N(c 2!2!2! M), we view the 2 M s as a single letter, so we now have a 10 letter word with two pairs of repeated letters. This gives N(C M ) = 10!, and similarly for 2!2! the other letters, so S 1 = 3(10!). Similarly, S 2!2! 2 = 3(9!), and S 2! 3 = 8!. By inclusionexclusion, the answer is S 0 S 1 + S 2 S 3 = 11! 2!2!2! 3(10!) + 3(9!) 8!. 2!2! 2! 10. Find the number of ways of arranging the letters in MAT HEMAT ICS so that exactly two pairs of consecutive letters occurs (this uses Theorem 8.2 if necessary that theorem will be on your exam). 3
4 Answer: We re looking for E 2, which by the formula with m = 2 is S 2 3S 3 = 3(9!) 2! 38!. 11. Determine the generating function for the sequence 1, 2, 2, 8, 16, 32, 64,... where the formula a n = 2 n holds for every term except the third one. Answer: If the sequence was 1, 2, 4, 8, 16, 32, 64,..., the generating function would be 1 + 2x + (2x) 2 + = 1. Since a 1 2x 2 is 2 instead of 4, we subtract 2x 2, giving f(x) = 1 1 2x 2x Find a generating function for the number of ways of distributing n pennies among five children such that every child gets at least 2 cents and the youngest child gets at most 5 cents. Answer: For each child but the youngest, the generating function for the number of ways a child can get n pennies is x 2 +x 3 +x 4 + = x2. For the youngest child, it is 1 x x2 +x 3 +x 4 +x 5. We multiply (since we re splitting the coins up between children), so that gives us ( x 2 ) 4 (x 2 + x 3 + x 4 + x 5 ) 1 x Comment: As the problem was stated, you could have left (x 2 +x 3 +x ) unsimplified. If I had said in closed form, then you would have had to use the geometric series formula to simplify it down to x2 1 x. 13. Find the x 5 coefficient of x 1 3x. Answer: This is the same as the x 4 1 coefficient of, which by the binomial theorem is 1 3x ( ) ( ) ( 3) 4 = ( 3) 4 ( 1) 4 = Comment: In this case you could also have used the geometric series formula. But if I x had instead asked about, you would have had to have used the binomial theorem. (1 3x) Find the constant (the x 0 coefficient) of (3x 2 x 1 ) 15 Answer: ( When we expand this out by the binomial theorem, the terms will look like 15 ) k (3x 2 ) k ( x 1 ) 15 k for various k. The x cancels for k = 5, so this is the term we want. The answer is ( ) ( 1) 10 = 3 5( )
5 15. Write down, but do not solve, a recurrence relation for the number of ways of writing n as a sum of 2 s, 3 s, and 5 s. Assume that order matters, that is that and are different. Answer: Let a n be the number of ways of writing n as a sum of 2 s, 3 s, and 5 s. If we consider the last number in our sum, it is either 2, 3,or 5. If it is 2, the remaining numbers have to add up to n 2, so there are a n 2 sums ending in 2. Similarly, there are a n 3 sums ending in 3 and a n 5 sums ending in 5. We therefore have a n = a n 2 + a n 3 + a n 5. Since this is a 5 th order recurrence, we need five initial conditions. We can see by inspection that a 1 = 0, a 2 = 1, a 3 = 1, a 4 = 1, a 5 = 3. Comments: Don t forget the initial conditions! 16. Solve: a n 3a n 1 + 2a n 2 = 2 n, a 0 = 0, a 1 = 1. Answer: We first solve a n 3a n 1 +2a n 2 = 0. The characteristic equation is r 2 3r+2 = 0, with roots 1 and 2, so the solution is A + B2 n. Next, we guess a solution to a n 3a n 1 + 2a n 2 = 2 n. Say we tried to guess C2 n. In this case a n 3a n 1 + 2a n 2 = C2 n 3C2 n 1 + 2C2 n 2 = C2 n ( ) = 0, which is hard to make equal to 2 n. So instead we go back and guess Cn2 n. This gives a n 3a n 1 +2a n 2 = Cn2 n 3C(n 1)2 n 1 +2C(n 2)2 n 2 = C2 n (n 3 2 (n 1)+1 2 (n 2)) = C2n ( 1 2. We want this to be equal to 2 n, so we set C = 2. Next we add our two solutions, getting 2n2 n + A + B2 n. Finally, we plug in our initial conditions. a 0 = 0 gives A+B = 0. a 1 = 1 gives 4+A+2B = 1. Solving gives B = 3, A = 3. So our final solution is 3 + (2n 3)2 n. We can check our solution by plugging in 2. Our formula gives a 2 = 7, and a 2 3a 1 +2a n 2 = 7 3 = 4 = 2 2, so it all checks. Comments: The problem we had initially when our guess turned the left hand side into 0 as a sign that we needed to multiply our guess by n. We could have known in advance this would happen, since our guess C2 n was also a solution to the homogenous recurrence. 17. Draw a 3 regular graph on 25 vertices, or explain why no such graph exists. Answer: No such graph exists. If it did, we would have 2e = deg(v) = (25)(3) = 75, so e would not be an integer. 5
6 18. Determine whether the graphs in figure in page 537 of your book are isomorphic or not. Answer: The graphs are not isomorphic. Acceptable reasons would include (i) G 2 contains a 4 cycle where every vertex has degree 4 (wztu). In G 1, the vertices of degree 4 are c, b, f, g, and do not make a cycle. (ii) In G 1, every vertex is contained in a 3 cycle. In G 2, y is not contained in any such cycle. (iii) Graph G 2 contains three 3 cycles intersecting in a single point. Graph G For which n does the Hypercube Q n have an Eulerian trail? Answer: Q n has 2 n vertices, each having degree n. If n is even, every vertex has even degree, so there is an Eulerian circuit. If n is odd, all the vertices have odd degree, so there is only a trail if n = Is graph (f) on page 554 of your book planar? Why or why not? Answer: The graph is not planar. It contains K 5, as can be seen from the below drawing. 6
7 Comment: How did I find this graph? I knew that if I wanted to find a K 5 somewhere inside the graph, the vertices I put in the K 5 had to have degree at least four. There were only five vertices like that in the original graph. Then I checked and saw which pairs of vertices of my five weren t connected in the original graph. For each of those pairs, I found a path going through one of the vertices I hadn t used so far in my graph. 21. (11.4, problem 19): Let G be a loopfree, connected, 4regular, planar graph with 16 edges. How many regions are in a planar drawing of G? Answer: By the formula deg(v) = 2e, we see that the sum of the degrees of the vertices must be 32. Since each vertex has degree 4, there must be 8 vertices. Now we use v e+r = 2. We know v = 8 and e = 16, so r = politicians get together at a party. It is known that each politician despises exactly 7 of his colleagues at the party. Show that we can put all 16 politicians in a receiving line so that nobody has to stand next so someone he or she despises. Answer: Draw a graph where each politician is connected to the people he or she does not despise, not counting him or herself. By assumption, each person is connected to 8 other people, and the graph has 16 vertices. It follows that the graph has a Hamiltonian cycle. We start at an arbitrary point in the cycle and just follow the cycle to get our receiving line. 23. A class of 5 people are trying to divide up into teams to work on at most 3 different projects. It is known that (Alice and Bob), (Bob and Charlie), (Charlie and Dave), (Dave and Alice), (Alice and Edna), and (Edna and Bob) are all incompatible pairs...putting both people in a pair on the same project would be a disaster. a) Express this in terms of a graphcoloring problem. b) Use chromatic polynomials to find the number of different arrangements of teams possible. Answer: We consider a graph where we connect two people if they cannot work together. The situation is equivalent to trying to find a 3 coloring of the graph labelled G below. 7
8 We use the chromatic polynomial recurrence to write P (G, λ) = P (G e, λ) P (G e(λ)). We have P (G e, λ) = λ(λ 1) 3 (λ 2) (there are λ choices for E, λ 1 for A, λ 2 for B, and λ 1 for each of C and D). For P (G e, λ), we see that there are λ choices for A, then λ 1 for B, then λ 2 for each of C and D, so P (G e, λ) = λ(λ 1)(λ 2) 2. Plugging into our recurrence, we see that P (G, λ) = (λ)(λ 1) 3 (λ 2) (λ)(λ 1)(λ 2) 2 We want to use three colors, so λ = 3, and we get P (G, 3) = 24 6 = Prove that any graph on 10 vertices with at least 26 edges must have chromatic number at least 3. Answer: This is really one of your homework problems in disguise. Suppose the chromatic number was 2 or less. Then the graph would be bipartite. If there are k vertices in one class and (10 k) vertices in the other class, there are at most k(10 k) edges connecting the two classes. But this can never be larger than 25, and our graph has 26 edges. Therefore the graph must not be bipartite, and therefore must have chromatic number at least two. Comments: Alternatively, you could think of k as the number of vertices of one color and 10 k as the number of vertices of the other color. It s really the same thing. 25. Find the tree on 8 vertices with Prufer code Answer: Code Vertex Set Edge ,2,3,4,5,6,7, ,3,4,5,6,7, ,3,5,6,7, ,3,6,7, ,6,7, ,7, ,8 38 The tree is the one with the seven edges from the edge column. 26. Problem 3 from section 13.1 Sketch of Answer: The final labels end up being a = (0, )b = (5, a), c = (6, a), f = (12, c), g = (16, h), h = (12, b). 27. (based on 13.4, number 2): Cathy is liked by Albert, Joseph, and Robert; Janice by Joseph and Dennis; Theresa by Albert and Joseph; Nettie by Dennis, Joseph, and Frank; and 8
9 Karen by Albert, Joseph, and Robert. USING THE ALGORITHM FROM THE PROOF OF HALL S THEOREM, find a way of setting up all five women up on dates with men who like them. Be sure to show and explain your procedure. Answer: We start with a bipartite graph where a woman is connected to a man if the man likes them. Initially the entire graph is colored blue. We will take each woman in turn and add her to our matching. Cathy: Connect to Albert. Albert is unused, so stop. At this point the only red edge is (Cathy, Albert) Janice: Connect to Joseph. Joseph is unused, so stop. At this point the red edges are (Cathy, Albert) and (Janice, Joseph) Theresa: Connect to Albert. Albert is already matched to Cathy, so we move down the list and connect Cathy to Joseph. Joseph is already connected to Janice, so we move down her list and connect her to Dennis. This gives the alternating path Theresa Albert Cathy Joseph Janice Dennis. Swapping the colors of every edge on that path leaves (Theresa, Albert), (Cathy, Joseph), and (Janice, Dennis) as red. Nettie: We first connect her to Dennis, who is paired with Janice. Janice can be connected to Joseph, who is paired with Cathy. Cathy is connected to Robert, who is currently unused. Swapping edges on the alternating path Nettie Dennis Janice Joseph Cathy Robert gives the new red edges as (Theresa, Albert), (Cathy, Robert), (Janice, Joseph), (Nettie, Dennis). Karen: Karen likes Albert, who is paired with Theresa, who likes Joseph, who is paired with Janice, who likes Dennis, who is paired with Nettie, who likes Frank. Switching the edges on this path gives new red edges of (Karen, Albert), (Theresa, Joseph), (Janice, Dennis), (Nettie, Frank), (Cathy, Robert), and we stop. Comments: I could have sped things up somewhat if I had been a bit more careful in choosing my alternating paths, for example just connecting Nettie to Frank instead of looking for a long alternating path. Really what I would be looking for here is just an explanation that shows that you understand the gist of the algorithm (continually searching for and swapping alternating path), rather than that you arrive at the correct final matching. 28. (13.4, number 7): Fritz is in charge of assigning students to parttime jobs at the college where he works. He has 25 student applications, and there are 25 different parttime jobs available on the campus. Each applicant is qualified exactly four jobs, but each job can be performed by at exactly four applicants. Can Fritz assign all the students to jobs for which they are qualified? Explain. 9
10 Answer: Yes, he can. Consider a group of k students. If they all listed out the jobs they could do, there would be a total of 4k items on all the lists, but each job would only appear in at most 4 lists. This means, by the pigeonhole principle, that at least k different jobs must be appearing on the lists. In other words, any group of k students has k different jobs they can do. By Hall s theorem, there must be a perfect matching of students to qualified jobs. Comment: This is just the cards in columns problem in disguise. 10
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