Illinois State University Mathematics Department

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

## Notes on In-Class Explorations Session #3

• Chapter 3 Section 2: pp 55-56, #10, #13, #18, #19

Chapter 3 Section 2
• #10: Begin by selecting a color for the middle horizontal strip of the flag.There are 5 ways to do this. For each of these 5, consider the adjacent horizontal stripes. There are 4 remaining color options for each of these stripes, in order to meet restriction (ii) that no two adjacent stripes are the same color. By the multiplication principle, for each uniquely colored center stripe, there are 16 possibilities. Here is an illustration using colors A,B, C, D, and E:
 B A B B A C B A D B A E C A B C A C C A D C A E D A B D A C D A D D A E E A B E A C E A D E A E

Notice, though, that, based on restriction (iii), some of these are duplicates, such as the two highlighted above [B-A-C and C-A-B]. We need to account for these duplicates. They are highlighted above using cell background color. Thus, we need to remove 6, leaving us with 10 potential flags when we fix one color for the center stripe. Note that we do not simply divide 16 by 2, because the diagonal entries (first and last stripe same color) do not have a duplicate.

We can repeat this with each of the four remaining center-stripe colors. This gives us 10 * 5 = 50 possible flags under the restrictions and conditions stated in the problem.

• #13: Because no team can have both 3 people on it and 4 people on it at the same time, the two situations are distinct. We compute the possibilities for each and add the result. We use combinations because there is no information telling us to distinguish one team member from another. Therefore, the total number of unique teams that are possible is C(17,3) + C(17,4).
• #18: We have to have a 40 years * 365.25 days/year (leap year inclusion) = 14,610 3-joke sets, with the order the jokes are told not distinguishing one set of three from another. This leads to the need to determine the value of n such that C(n,3) >= 14,610. You can substitute C(n,3) = (n!)/[(n-3)!3!] and solve the resulting cubic inequality or you can use a educated-guess-and-check strategy to determine n. Here, we smallest integer value of n that satisfies the inequality is n = 46. Therefore, Juanita must have at least 46 jokes in her joke set.
• #19: (a) We need to consider distinct cases in order to avoid overcounting. To assure at least 3 faculty and 3 students, we have three distinct cases:
• Case I: exactly 3 faculty and 5 students. We can have C(20,3) * C(20,5) different committees of this make up.
• Case II: exactly 4 faculty and 4 students. We can have C(20,4) * C(20,4) different committees of this make up.
• Case I: exactly 5 faculty and 3 students. We can have C(20,5) * C(20,3) different committees of this make up.

Because each case has no overlap with other cases, we add to determine the total number of committees that are possible: C(20,3) * C(20,5) + C(20,4) * C(20,4) + C(20,5) * C(20,3) = 2[C(20,3) * C(20,5)] + [C(20,4)]2.
Note that C(20,3)*C(20,3)*C(34,2) is not a correct solution nor does it illustrate a correct strategy!

(b) Here, it is more efficient to determine the total number of committees possible and remove those with no faculty: C(40,8) - [C(20,0) * C(20,8)].