
Chapter 2 Section 2 (pp 2627)
 #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 3233)
 #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 7digit 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 3336)
 #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 4247)
 #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(n1)...(5)
 #11: Oliveras claim is equivalent to claiming that n!/(nr)! = n!/(r1)!. For this to be true, we would need (nr)! = (r1)!, or, equivalently, nr = r1. 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 5156)
 #1: C(20,6) = 20!/(6!14!) = (1/6!)*(20!/14!) = P(20,6)/6!
 #3: (a) 14*13*12*11 (b) 14!/10! (c) 4!*C(14,4)
 #5: r=23
 #9: C(750,10)
 #11: (a) C(48,20) ["selecting" not "arranging"] (b) C(26,20) + C(42,20)
 #14: C(8,4)*C(13,5)
 #15: (a) P(8,8) (b) 6! (c) 4*3*6! (d) C(8,5)
 #20: 46 jokes: Determine j so that C(j,3) >= (365)(40)+11 [addition of 11 covers most possible leap years that could occur]
 #22: See solution to Problem 6, Test 2, 1996 course: http://www.math.ilstu.edu/day/courses/old/305/old/96/Test02S96soln.html
 #24: See solution to Problem 5 of Test #3, 2001 course: http://www.math.ilstu.edu/day/courses/old/305/old/01/test03soln.html#5

