Illinois State University Mathematics Department

 MAT 409: Topics in Algebra and Combinatorics for K-8 Teachers

Notes on Assignment #10

 Assignment #11 Read Chapter 5 Chapter 6 Chapter 7 Problems Chapter 4 (pp 128-130): 16 Chapter 5 Section 1 (pp 151-155): 5 Chapter 5 Section 2 (p 161): 1,2 Chapter 6 (pp 172-177): 2,5,8,11,15,16,17,20,32

Chapter 4
• #16: In a standard deck of 52 cards, there are 13 cards of each of four suits. To collect a set of four cards each of different suits, we must select one card from each suit. This can be done in 13×13×13×13=134 ways.
Chapter 5 Section 1
• #5: (a) and (b): Because "no family was completely without such items," we know all 983 families had at least one of the items. We now compare information about the number of families with known numbers of items. Because 345 families had at least 3 items, and 264 families had all 4 items, the difference, 81 families tells us the number of families with exactly 3 items. Using this same process, we can determine that 136 families possessed exactly 2 items, and 502 families possessed exactly 1 item. Note that 502+136+81+264 = 983. Use this four values to determine the total number of items held by all 983 families:
• 502×1 + 136×2 + 81×3 + 264×4 = 2073 items
Chapter 5 Section 2
• #1: This is the number of derangements of n items: D(n) = n!(1 - 1/1! + 1/2! - ... + (-1)n/n!)
• #2: D(6) = 265
Chapter 6
• #2: s(0)=5, s(1)=2(5)+3=13, s(2)=32, s(3)=73, s(4)=158, s(5)=331
• #5: s(0)=1, s(1)=2, s(2)=5, s(3)=5+2(2)+1+3=13, s(4)=29, s(5)=65
• #8: The word "sequence" implies that order matters.
• \$1: either Tape (T) or Ruler (R), so S(1)=2
• \$2: TT, RR, TR, RT, or Pen (P1), Pencil (P2), Paper (P3), so S(2)=7. Note that this is 2·S(1)+3.
• \$3: To each \$1 purchase, we can attach one of three \$2 items; to each \$2 purchase, we can one of two \$1 items; we can also purchase a Binder (B). So S(3) = 21. Note that this is 3·S(1)+2·S(2)+1.
• \$4: 2·S(3)+3·S(2)+S(1)=65
• So for \$N, we have the recurrence relation S(N) = 2·S(N-1) + 3·S(N-2) + S(N-3), with initial conditions that S(0) = 1, S(1) = 2, and S(2) = 7.
• #11: From previous activities, we know that there are 2n subsets for a set with n elements. So for n = 1,2,3,..., we have s(n) = 2,4,8,.... Recursively, this means s(n) = 2·s(n-1) with s(1) = 2.
• #15:The word "select" indicates that the order items are chosen is not significant. Given the kind and number of fruits available, the fruit pairs include AA, AP, AO, AB, PO, PB, OB. This is a collection 7 fruit pairs.
• #16: We assume here that each person wants one fruit danish, and that we are not concerned about which person ordered which danish, merely that orders were placed.
• 0 people ordering: 1 way. d(0) = 1
• 1 person ordering: 3 choices, so there are three possible orders. d(1) = 3
• 2 people: Because there are at least two of every danish, we can consider all ways to collect two danish: CC,AA,RR,CA,CR,AR. d(2) = 6.
• 3 people: Now we must be concerned not to sell more than 2 apricot danish among the possible orders: CCC,RRR,CCA,CCR,AAC,AAR,RRC,RRA,CAR. d(3) = 9.
• 4 people: Now we must be concerned not to sell more than 3 cheese danish as well as only 2 apricot danish among the possible orders: RRRR,CCCR,CCCA,RRRC,RRRA,RRCC,RRAA,CCAA,RRCA,CCAR,AACR. d(4) = 11.
• Now we begin to use the symmetry for set inclusion/exclusion. For instance, because only 9 danish are available, The 11 orders possible for 4 people, d(4) = 11, means 5 danish are not ordered. Thus, d(5) = 11, d(6) = 9, d(7) = 6, d(8) = 3, and d(9) = 1.
• #17: In determining whether order is significant, consider the situation. If I spend \$7, I can get two cheese sandwiches and one bowl of soup. Does the order one arranges those items (CCS, CSC, SCC) make this three distinct orders? No, the order is two cheese sandwiches and one bowl of soup. So, here, the arrangement of the items within a food order is not significant. By brute force enumeration, we have:
• d(0) = d(1) = 0: There is no way to place an order whose value is \$0 or \$1.
• d(2) = 1
• d(3) = 1
• d(4) = 1
• d(5) = 1
• d(6) = 2
• d(7) = 1
• d(8) = 2
• d(9) = 2
• d(10) = 2
• d(11) = 2
• d(12) = 3
• #20: For the sequence 1,2,4,8,16, . . . , each new term is twice the previous: s(n) = 2·s(n-1), with s(1) = 1
• #32: We seek a recursive relationship so that the number of rectangles measuring 3-by-n can be described in terms of the number of rectangles of smaller length. We observe that there is one way to create a 3-by-1 arrangement, two ways to create a 3-by-2 arrangement and six ways to create a 3-by-3 arrangement:

• Now we can create new 3-by-n arrangements using these. For the 3-by-4 arrangements, first use the 3-by-1 arrangement and attach to it each of the last (bottom) three 3-by-3 arrangements shown above. Next, use the each existing 3-by-2 arrangement and attach a 3-by-2 tile. Finally, to each of the six 3-by-3 arrangments, attach a 3-by-1 tile. So, A(4) = 3·A(1) + A(2) + A(3).
• This recursion can be generalized to A(n) = 3·A(n-3) + A(n-2) + A(n-1), with A(1) = 1, A(2) = 2, and A(3) = 6.