Illinois State University Mathematics Department

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

## Notes on Assignment #9

 Assignment #9 Read Chapter 4 Section 1 Chapter 4 Section 2 Problems Chapter 4 (pp 128-130): 20,22 Chapter 5 Section 1 (pp 151-155): 1,2,4,7

Chapter 4
• #20: Let K represent the total number of letters in the word, and use Li to represent the number of repetitions of each letter, 1 <=  i <= 3, i an integer, such that L1 + L2 + L3 = K. We need, then, all unique arrangements of the letters such that K!/(L1!L2!L3!) = (L1 + L2 + L3)!/(L1!L2!L3!) <= 1000.
• #22: We assume here that each of the 26 letters occurs once. The letters c and d represent two of the 26 letters. There are 2! = 2 ways to arrange these two letters. Now we select 6 of the remaining 24 letters and permute them between c and d. This can be done in C(24,6)×P(6,6) = P(24,6) ways. This gives us a bundle of 6 letters (none to be c or d) surrounded by c and d. Now we permute the remaining 18 letters in P(18,18) = 18! ways. Within any arrangement of these 18 letters, there are 19 slots into which we can place the 8-letter bundle described previously. We choose one of those 19 slots, in C(19,1) = 19 ways. In all, then, there are P(2,2)×P(24,6)×P(18,18)×19 desired arrangements. Another way to consider this is to first permute the 24 letters no including c or d. This can be done in P(24,24) = 24! ways. Now, we count the number of slot-pairs we can place c and d such that they surround exactly 6 letters. In the line up of 24 letters, there are 25 slots. Let us call those slots 0 through 24. A slot-pair must have 6 letters between them, so the pair 0/6 is the first available slot-pair, continuing through slot-pair 1/7, 2/8, and so on, through 18/24. There are 19 different slot-pairs to choose from in placing the letter-pair c/d. We can use d/c as well. Therefor, there are 24!×19×2 total unique arrangements. The two expressions we have generated can be shown to be equivalent.

Chapter 5 Section 1

• #1: One way to solve this problem is to create a three-circle overlapping Venn diagram and start from the deepest overlap (all three magazines) and work your way out. Remember to include the 8 students who subscribe to none of these magazines. Now count all eight components of your diagram. You should get 99 in all. Another way to solve this problem is to solve use the Inclusion/Exclusion principle and solve for T, the total number of elements in a set. Here, we first generate the I/E equation 8 = T - (47 + 44 + 32) + (11 + 12 + 12) - 3. Solving for T we get T = 99.
• #2: (a) There are 24 = 16 different descriptions possible for a diner, because each diner is characterized using 4 categories, and each diner either has or does not have each categorical characterization. (b) We seek the number of non-negative integer solutions to the equation c1 + c2 + . . . + c16 = 38. We have 16 categories and 38 objects to place within them, so there are C(16 - 1 + 38, 16 -1) = C (16 - 1 + 38, 38) different ways to fill the categories with the 38 objects.
• #4: Use either the I/E principle or a Venn diagram to generate data that sets up a contradiction. Using the I/E principle, we have the equation 111 = 144 - (16 + 13 + 17) + (10 + 9 + X) - 5. Solving for X, which represents the number of people who had been to both Europe and Asia, we get X = -1, which is impossible.
• #7: By the I/E principle, we solve the following inequality for T: 1 <= T - (15 + 14 + 18) + (8  7 + 9) - 6. This yields 30 <= T. So, the smallest total enrollment is 30.