Illinois State University Mathematics Department

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

## Notes on Assignment #3

Chapter 3 Section 2
• #3: P(14,4) as (a) product of 4 consecutive integers: 14 × 13 × 12 × 11; (b) using factorial notation: 14!/(10!); (c) in terms of C(14,4): C(14,4)*4!
• #5: C(42,19) = C(42,42-19) = C(42,23), so r = 23
• #7:

• #11: (a) There are no restrictions about type of token nor are we concerned about order the token are chosen, so we pool the tokens and choose 20: C(68,20). (b) Now we only select from those of the same color. There are C(26,20) ways to select 20 token all red and C(42,20) to select 20 green tokens. Because these are disjoint selections and we are choosing one token, we add these results. This yields C(26,20) + C(42,20) ways to select 20 tokens all of one color.
• #15: There are P(14,14) = 14! ways to arrange the 14 letters in the word, but some of those arrangements will be duplicates, because we have some letters appearing more than once. For instance, if we label the two L letters as L1 and L2, we have these two arrangements among the 14! that are possible: QUADRIL1L2IONTHS and QUADRIL2L1IONTHS. Because the letters L are identical, these two arrangements are the same. To account for this duplication, we divide 14! by the number of ways each of the duplicated letters can be permuted. This gives u 14!/(2!2!) total unique arrangements.
• #18: We need to have 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 an educated-guess-and-check strategy to determine n. Here, the 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 over-counting. 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)].

Chapter 3 Section 3

• #1: C(7,k), with 0 ≤ k ≤ 7
• #7: (a) We need to traverse 8 blocks southward and 12 blocks eastward. This yields C(20,8) = C(20,12) 20-block paths from T to B. (b) One way to solve this problem is to consider the number of possible paths with no restrictions, C(20,8) = C(20,12), then count and remove those that can't be traversed. In the scenario described here, it is a 9-block drive to the west end of the street to be closed. There are C(9,4) = C(9,5) ways to get to that point. There is only one path from that point to the east end of that restricted street: C(1,1). Finally, there are C(10,4) = C(10,6) paths from that east-end point to point B. We multiply these three counts together, to match them: C(9,5) × C(1,1) × C(10,6). We then subtract this number of closed paths from the original total number of paths: C(20,12) − [C(9,5) × C(1,1) × C(10,6)].
• #11: This is analogous to the birth-order problems we explored: C(6,4) = C(6,2).
• #17: 12!/(4!2!2!)
• #26: (a) 21!/(3!2!2!2!2!2!2!3!) (b) There are 11 ways to choose the first (left-most) consonant and 10 ways to choose the last (right-most) consonant. There remain 19 letters to permute, to be placed between these two consonants. We still need to account for all the duplication of arrangements due to repetition of letters. This yields (11×10×19!)/(3!2!2!2!2!2!2!3!) different such arrangements. (c) Concentrating on the vowels, we have 10!/(3!3!!2!2!) different arrangements of the vowels. We now have 11 consonants to arrange, yielding 11!/(2!2!!2!2!) different consonant arrangements. Because there are no restrictions on how the consonants are positioned (relative to consonants only), we can think of the 11 consonants as borders and the 12 spaces around the consonants as positions within which the one vowel group could be placed. This yields 12 spots into which we could drop the vowel group. We now match each consonant arrangement with each vowel arrangement, using the Multiplication Principle, and, finally, multiply that by 12: 12 × 10!/(3!3!!2!2!) × 11!/(2!2!!2!2!) different arrangements. (d) There are 11 unique letters in the word, and we select 5 of them, with no regard for order: C(11,5).