Illinois State University Mathematics Department

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

Notes on Assignment #4

  • Chapter 2 Review (pp. 35-38): 5,6
  • Chapter 3 Section 1 (p. 47): 14
  • Chapter 3 Section 2 (p. 56): 17
  • Chapter 3 Section 3 (pp 70-77): 3,5,10,18,31,33
  • Chapter 3 Section 4 )pp 94-98): 2,6,10,16,21,32

  • Chapter 2 Review
    • #5: Using the multiplication principle, there are 6∙3∙8 = 144 meal combos. By the Pigeonhole Principle, there must be 144 + 1 = 145 meal orders placed to assure that at least one meal combo is ordered twice.
    • #6: Using the addition principle, there are 5 + 7 + 4 = 16 different breakfast sandwiches.
    Chapter 3 Section 1
    • #14: With no restrictions whatsoever, there are 26∙26 two-letter arrangements. There are restrictions, however. There are 26 two-letter arrangements with duplicate letters. There are 25 adjacent pairs, from ab and bc to xy and yz. We double 25 because the order of adjacency isn't restricted, so ba, cb, . . . , yx, zy also cannot be used. So from 26∙26 we remove (26 + 2∙25): 26∙26 - 26 − 2∙25 = 26∙(26−1) − 2∙25 = 26(25)−2(25) = 24(25) = 600.
    Chapter 3 Section 2
    • #17: For the set of digits {0,1,2,3,4,5,6,7,8,9}, each digit either is in a subset or it is not. Thus, for each digit, there are two options for subset inclusion. By the multiplication principle, there are 210 different subsets. We must exclude the empty set and we must exclude the subset that contains all ten digits. So we have 210 − 2 = 2∙(29-1) different allowable subsets.
    Chapter 3 Section 3
    • #3: (a) Position the 13 non-distinguishable mathematics (M) books. Among these books, there are 14 places to position a science (S) book so it will not be adjacent to other S books. So from those 14 spots we choose 7: C(14,7). (b) If the books can be distinguished, there are 13! ways to first arrange the M books. Then, there are P(14,7) ways to arrange the S books among these M books so that no two S books are adjacent. We cross-match these options, using the multiplication principle, to get 13!∙P(14,7) unique shelving arrangements.
    • #5: (a) The path is 19 blocks long. Imagine a 19-spot arrangement within which we position 12 identical characters W and 7 identical characters S. There are C(19,12) ways to select spots for the 12 identical W's. Once positioned, there is only one way, C(7,7), to position the identical S's. Thus, there are C(19,12) paths for Geovanna. (b) This restriction requires Geovanna to walk S first. From there, she can walk any 18-block path that goes 6 blocks S and 12 blocks W. The number of such paths, justified using the same argument as in (a), is C(18,12). (c) The only way Geovanna can walk a 21-block path and go from home to the library is to either (i) include one additional S block then account for it with a N block, or, (ii) include one additional W block and account for it with an E block. Option (i) results in C(21,8)∙C(13,12)∙C(1,1) and option (ii) results in C(21,13)∙C(8,7)∙C(1,1). Because these two options result in disjoint paths, there are C(21,8)∙C(13,12)∙C(1,1) + C(21,13)∙C(8,7)∙C(1,1) such paths in all.
    • #10: This is similar to the question posed in class about birth-order arrangements, by gender, in a family. We have 10 flips in all, with 6 identical HEAD (H) results and 4 identical TAIL (T) results. There are C(10,6) = C(10,4) arrangements. If k of the flips are T, we have C(10,k) = C(10,10-k) arrangements.
    • #18: In the word PREFERRER, there is one P, four letters R, three appearances of E, and one F, making 9 letters in all, four of which are distinct. This is an arrangement when repetition is allowed. We have 9!/(1!4!2!1!) possible arrangements. This is the same as C(9,4)∙C(5,3)∙C(2,1)∙C(1,1).
    • #31: We first pair up the paperbacks, place them, and then insert hardback books between the pairs to assure non-adjacency of the pairs. To complete the first task, we can think of grabbing two paperback books at a time and arranging them. This can be done in C(6,2)*P(2,2) * C(4,2)*P(2,2) * C(2,2)*P(2,2) ways. Simplifying this expressions, we find it is equivalent to P(6,6) = 6!. Thus, we could think of completing the first task as simply arranging the six paperback books. We can consider arrangement of the three hardback books in the same way. There are P(3,3) = 3! ways to line up those books, regardless of where they go. Now we must place them. What are the possibilities for these three hardbacks? Let us use PB to represent two paperback books together and H to represent one hardback. We could have: H-PB-H-PB-H-PB or PB-H-PB-H-PB-H or PB-HH-PB-H-PB or PB-H-PB-HH-PB. Without changing the order of the set of three hardbacks, we have four ways to position the H books and still meet pair-adjacency requirements of the PB books. Remember, again, that P(3,3)=3! had previously accounted for all linear arrangements of the three HB books, so this last sequence of counts was just about choosing spots to assure the PB pairs restrictions. All of these counts are match-ups, so we multiply: 6!∙3!∙4.
    • #33: Here we need to look at disjoint cases determined by the number of games in a series. There are four such cases, whose count results we will add at the end. For each case, it's significant to note that the series must end with a W. That is, for the winning team, that team's last game must have been a win.
      • Case I: 4-game sweep. There is only one way for this to be accomplished: WWWW. In the format of the following cases, that is C(3,3), with a W required in 4th position.
      • Case II: a 5-game series. We must place a W in the fifth position. Now we arrange the remaining three W's among the first four positions: C(4,3). With that task complete, the single L is already determined.
      • Case III: a 6-game series: Again, we need a W in the last position (game six). Now we arrange the remaining three W's among the first five positions: C(5,3). With that task complete, the two L's are already determined.
      • Case IV: a 7-game series: Again, we need a W in the last position (game seven). Now we arrange the remaining three W's among the first six positions: C(6,3). With that task complete, the three L's are already determined.
      • Now we add the counts from the four disjoint cases: C(3,3) + C(4,3) + C(5,3) + C(6,3).
    Chapter 3 Section 4
    • #2: C(18,9) = C(17,9) + C(17,8)
    • #6: We know C(10,5) = C(9,5) + C(9,4) using Pascal's Formula. We manipulate this to get C(10,5) - C(9,5) = C(9,4). The desired single combination expression is C(9,4).
    • #10: In (2a + 5b)n there are n + 1 collected terms.
    • #16: In the expansion of (a + b + c)8, we seek the number of different arrangements of the eight characters aaaabbbc. There are 8!/(4!3!1!) such unique arrangements.
    • #21: (a) In (2a + b)5 there are n + 1 collected terms, with n = 5, so there are 6 collected terms. (b) The expansion and collection begins with C(5,5)*(2a)5*b0 + C(5,4)*(2a)4*b1 + C(5,3)*(2a)3*b2 =  1*32a5b0 + 5*16a4b1 + 10*8a3b2 = 32a5 + 80a4b + 80a3b2
    • #32: In the expansion of (a + b + c + d)8, we seek the number of different arrangements of the eight characters aabbbbcd. There are 8!/(2!4!1!1!) such unique arrangements.