Illinois State University Mathematics Department

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

## Notes on Assignment #2

Chapter 2 Section 2 (pp 28-29)
• #2: disjoint sets: I and II; I and III; I and IV; II and III; II and IV; III and IV; IV and VI
• #3: I, II, III; I, II, IV; I, III, IV; II, III, IV; I, II, III, IV
• #4: {f,j,k,l,q,v,w,z}
• #5: 10 choices
• #6: 8 choices
• #11: 17 (5+6+2+4)
Chapter 2 Section 3 (pp 34-35)
• #2: 5 * 6 * 2 * 4 = 240 ways
• #3: Look at the most restrictive portion first: the units digit of the number. It must be odd, so there are 5 choices for that position. Because the digits in the number must be unique, there remain 9 digits to choose from. We fill the remaining three spots (tens, hundreds, thousands positions) in 9 * 8 * 7 = 504 ways. So, in all there are 5 * (9 * 8 * 7) = 2520 integers.
• #5: Begin by determining how many 7-digit numbers there are that have unique digits and no 0, then remove from this total those that contain 5 and 6 adjacent. There are 9*8*7*6*5*4*3 7-digit numbers that have unique digits and no 0. Of these, we cannot have 5 and 6 adjacent. The digits 5 and 6 can be adjacent as ...56... and as ...65.... Also, within any 7-digit number, a 56 package can appear in any one of six positions in the number (first/second, second/third, and so on). So we have 2*6 = 12 ways/places 5 and 6 may appear adjacent. With each of these 12 ways, however, we must consider options for all the other digits in the 7-digit number. We have already used 5 and 6, so we have 7 available digits with which to build the rest of the number, containing 5 open spots for digits. We can fill those spots in 7*6*5*4*3 ways/ Therefore, the total restricted 7-digit numbers is (12) * (7*6*5*4*3). Therefore, the number of desired 7-digit numbers is (9*8*7*6*5*4*3) - [(12) * (7*6*5*4*3] = 151,200.
• #7: Each person can choose from 5 shirts. The desired number of 3-shirt sets is 5*5*5 = 125.
• #8: Consider the 5 consonants. There are 5! = 5*4*3*2*1 ways to arrange these. With each arrangement of the consonants, we have six places into which we can drop vowels and meet the requirement of vowel non-adjacency:
___ C___M___P___T___R___
We choose three of these six available spots and permute the vowels there. There are C(6,3) * 3! = P(6,3) such ways. In all, then, there are (5!) * (P(6,3)) = 14,400 ways.
Chapter 2 Review (pp 35-38)
• #2: 96 responses
• Because there are five categories of STEM professionals, The surveyor could get 19 responses from each of the five groups and still not have 20 from a group. When the surveyor gets that 96th response, however, it will assure there are at least 20 responses from the same group.
• #4:
• (a) 20 choices, by the Addition Principle.
• (b) 96 pairs: using the multiplication principle, the pair each of the 8 coffee choices with each of the 12 tea choices. We disregard order, as stated in the problem
• #7:
• (a) There are two choices for each of the six dots, either raised or not raised, so there are 26 = 64 different raised-dots options. One of those, however, is no raised dots, which is not an option. Therefore, there are 63 symbols that can be represented.
• (b) Even numbers are 0, 2, 4, 6, and so on. Here, 0 raised dots is not an option. So for 2, 4, and 6 raised dots, we use combinations to determine the number of ways to select dots for raising. We have C(6,2), C(6,4), and C(6,6). Because these are disjoint (we cannot have exactly two raised dots and at the same time have exactly four raised dots), we add these results: C(6,2) + C(6,4) + C(6,6) = 31 symbols with an even number of raised dots.
• (c) There are at least a couple ways to do this.
• The first method is to look at options that meet the restrictions. In each row, there must be one or no raised symbols. This can be done in 3 ways for each row: only left dot raised, only right dot raised, or neither dot raised. Because there are three rows with this required characteristic, there are 3*3*3 = 27 different options. But one of these options--no dot raised in any row--isn't allowed, so there are 26 symbols, in all that are possible under this restriction.
• A second method is to start with all possible symbols (63, from part (a) here) are remove those that cannot be used. Here, those that cannot be used include any circumstances with both dots raised in one or more rows. We need to look at cases here.
• We could have 2 dots raised in every row. There is only one way to do that.
• We could have two dots raised in two of the three rows. There are 3 ways [C(3,2)] to choose exactly two rows at a time, and for each of these options, there is one way to raise two dots in each of the two rows and three ways to raised less than two dots in the remaining row. Thus, for each two-row pair here, there are 1*1*3 = 3 symbols, and, as stated, there are three 2-row options for this. Thus, There are 9 symbols for this case.
• The last case is two raised dots in exactly one row. Using similar reasoning as the previous case, there is one way to raise two dots in that required row and three ways to choose less than two dots in the other two rows. Each of these is possible for each of the three ways we can select one row for the two dots. So, for this last case, there are (1*3*3)*C(3,1) = 9*3 = 27 symbols.
• Add the three cases (disjoint outcomes), there are 1 + 9 + 27 = 37 ways to have one or more rows with two raised dots. We subtract 37 from 63 to show, once again, that there are 26 symbols with no more than one raised dot in any row.
Chapter 3 Section 1 (pp 45-50)
• #1: 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
• #2: n! = n*(n-1)*(n-2)*...*2*1
• #6: 10!/(8!2!) = (10*9)/2 = 45
• #8: n!/4! = n*(n-1)*(n-2)*...*6*5
• #13: (a) P(8,5) = 6720 (b) 85 = 32,768
• #15: (a) 6! * 8! = 29,030,400 (b) (7*6) * (6*5) * 9! = 457,228,800 (c) 7! * 6! = 3,628,800
• #18: (a) 8! = 40,320 (b) 6! = 720 (c) (4*3) * 6! = 8640
• #23: (5*6*2*4) * 4! = 5760