Illinois State University Mathematics Department

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

 Syllabus Grades & Grading Session Notes Assignments and Problem Sets Tests and Quizzes

## Notes on Assignment #2

Chapter 2 Section 2 (pp 26-27)
• #2: 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; II/III/IV; I/II/III/IV

• #4: f,j,k,l,q,v,w,z

• #5: 10 letters

• #6: 8 letters

• #11: 17 choices

Chapter 2 Section 3 (pp 32-33)

• #1: 3*2*5=30

• #2: 5*6*2*4=240

• #3: 5*7*7*6; place units' digit first, then the thousands' digit, then the remaining two digits.

• #5: Consider all 7-digit values that meet the second and third condition. This can be done in P(9,7) ways. Now subtract those that have 5 and 6 adjacent in either order. To do this, first place five of the seven digits in {1,2,3,4,7,8,9}. Now, from among the six spaces existing among these five digits, place 5/6 together. This placement can be done in 6 ways. finally, consider that 5 and 6 can be reversed, so two ways exist for that. So we subtract P(7,5)*6*2. Final result: P(9,7) - P(7,5)*6*2. See also solution 7 of Supplement B, previous course (http://www.math.ilstu.edu/day/courses/old/305/supbsoln.html) for an alternative explanation.

• #6: P(8,5)*P(8,4)*P(7,5). Place 5 in front row, 4 in back row, and rest in remaining seats.

• #8: P(5,5)*P(6,3). Discussed in class.

• #10: 5*5*5

Chapter 2 Review (pp 33-36)

• #1: See strategy used in Assignment #1, Problem #16.

• #2: 96 responses

• #4: (a) 20 choices (b) 8*12=96 choices

• #8: Consider two cases: I: four characters: 2*36*36*36; II: five characters: 2*36*36*36*36. By addition principle, total choices is the sum (2*36*36*36) + (2*36*36*36*36)

Chapter 3 Section 1 (pp 42-47)

• #1: The value 6! is 720. The positive integer factors of 720 are 1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,30,36,40,45,48,60,72,80,90,120,144,240,360, and 720.

• #5: 6*5=30

• #7: 8

• #8: n(n-1)...(5)

• #11: Oliveras claim is equivalent to claiming that n!/(n-r)! = n!/(r-1)!. For this to be true, we would need (n-r)! = (r-1)!, or, equivalently, n-r = r-1. Using algebra, we see this will be true when 2r=n+1. So her claim is sometimes true. For instance, P(11,6) = 11!/5! fits her claim. A counterexample, however, is if n=10 and r=2: P(10,2)=10!/8!, but her claim would be that this is the same as 10!/1!, which is not the case.

• #13: (a) P(8,5) (b) 8*8*8*8*8 = 8^5

• #15: (a) 8!*6! (b) P(7,2)*P(6,2)*P(9,9) (c) 7!*6!

• #19: (a) P(200,6) (b) 44*52*18*21*36*29 (c) P(148,6)

• #24: 5!*6*4!

Chapter 3 Section 2 (pp 51-56)