Illinois State University Mathematics Department

 MAT 305: Combinatorics Topics for K-8 Teachers Spring 2001 Dr. Roger Day (day@ilstu.edu)

 Possible Solutions: Test #2

Please write your solutions on one side only of each piece of paper you use, and please begin each solution on a new sheet of paper. You may use factorial notation as well as combination and permutation notation where appropriate (i.e., there is no need to expand 24!).

You are to work alone on this test. You may not use anyone else's work nor may you refer to any materials as you complete the test. You may ask me questions.

Evaluation Criteria

You may earn up to 10 points on each of questions 1 through 6. For each question:

• 6 points count toward a correct solution to the problem. I will evaluate the mathematics you use:
• Is it accurate and appropriate?
• Have you provided adequate justification?
• 4 points count toward how you express your solution. I will evaluate how you communicate your results:
• Is your solution clear and complete?
• Have you expressed logical connections among components of your solution?

1.

Respond to each of these questions. While you may show steps leading to your solution, you do not need to generate written explanations for questions (a) through (e) on this page. (2 points each)

(a) What value K satisfies the equation P(12,4) = KoC(12,4)?

 Solution: K=4!

(b) How many distinct arrangements exist for the letters in the word mammilliform?

 Solution: 12!/(4!2!2!)

(c) In the expansion of (a+c+e+g+i)^10, state:

(i) the number of uncollected terms

 Solution: 5^10

(ii) the coefficient R in the collected term Ra^3cg^2i^4.

 Solution: R=10!/(3!2!4!)

(d) Determine the number of collected terms in the expansion of (x+y)^6.

 Solution: 7

(e) Replace r and s in C(10,5) &emdash; C(9,5) = C(r,s) to correctly illustrate Pascal's Formula, a fundamental relationship that exists in Pascal's Triangle.

 Solution:C(9,4)

2.

Geovanna walks from her apartment to the library every day. Her apartment is 7 blocks north and 12 blocks east of the library. Geovanna walks along city streets that are laid out in a grid system.

(a) Given access to all the streets in the system, how many different 19-block routes could Geovanna use to get from home to the library? (3 points)

 Solution:C(19,7)

(b) The east/west street running outside Geovanna's apartment is under repair. She cannot walk west on that street for the first 3 blocks outside her apartment. Now how many different 19-block routes could Geovanna use to get from home to the library? (3 points)

 Solution: C(18,6) Geovanna must first walk one block south. She now has an 18-block route to walk, to include 6 blocks south and 12 blocks west.

(c) Again given access to all the streets in the system, how many different 21-block routes could Geovanna use to get from home to the library? (4 points)

 Solution: C(19,7)*20*4 For each of the C(19,7) routes Geovanna could follow from her home to the library, there are 20 intersection points of E/W with N/S streets. At each intersection point, she could go in one of 4 directions one block and return to the same point.

3.

Referring to the letters in the word ENTEROBACTERIACEAE, solve each of the following problems. Each problem is independent and separate from the others.

(a) How many arrangements are there for the letters in this word? (2 points)

 Solution: 18!/(5!2!2!3!2!)

(b) How many arrangements exist if each cannot begin nor end with E? (2 points)

 Solution: (P(13,2)*16!)/(5!2!2!3!2!) Remove the Es from consideration and place a non-E at the beginning and a non-E at the end of the 18 spaces for letters. This can be done in P(13,2) ways. Now return the Es to the letters remaining for placement. There are 16! Ways to place the remaining 16 letters. We divide by 5!2!2!3!2! to account for duplications of letters.

(c) How many arrangements can be made if no two consonants can be adjacent to each other? (3 points)

 Solution: (10!*P(11,8))/(5!2!2!3!2!) Arrange the 10 vowels first. This can be done in 10! ways (duplicates will be accounted for all at once when done). There are now 11 places among the vowels into which we can place consonants. The 8 consonants can be arranged in these 11 places in P(11,8) ways. We divide by 5!2!2!3!2! to account for duplications of letters.

(d) If the only distinction we can make is between vowels and consonants, how many arrangements can be made? (3 points)

 Solution: C(18,10) Among the 18 places for letters, select 10 places for vowels. The 8 remaining places get the consonants.

4.

Roger is taking a True/False test in his automobile mechanics class. There are 10 unique items on the test. Using a numbered answer sheet, students are requested to circle either True or False corresponding to each item.

(a) How many different responses could Roger submit on the 10-item answer sheet, assuming he responds either True or False to each item? (3 points)

 Solution: 2^10 There are two choices at each of the 10 answer-blank positions on the answer sheet.

(b) If we allow for the possibility that Roger might not circle either True or False on any or all items (i.e., he could leave items blank), how many different answer-sheet responses could he submit? (3 points)

 Solution: 3^10 There are now three choices at each of the 10 answer-blank positions on the answer sheet.

(c) Roger's friend Howie took the True/False test before Roger did. Howie told Roger there were 5 True items and 5 False items. If Roger heeded Howie's advice and circled 5 True and 5 False on the answer sheet, how many different answer-sheet responses could he submit? (4 points)

 Solution: C(10,5) Among the 10 answer blanks, select 5 and circle True. The 5 remaining blanks must be False.

5.

Consider the expansion of the binomial (2a-b)^12.

(a) How many uncollected terms are in this expansion? (4 points)

 Solution:2^12

(b) In the collected term Ha^5b^7, what is the numerical value of H? (6 points)

 Solution: H = C(12,7)*2^5*(-1)^7 = -25,344

6.

Choose one of the following problems and solve it in the space provided. If more than one solution appears, I will evaluate only the first one I encounter.

I. A shelf is to contain nine different books, six different paperback books and three different hardback books. If the paperback books must be shelved in pairs (that is, exactly two paperback books must be adjacent to each other), how many ways can the nine books be shelved?

 Solution: There are 3!*C(4,3)*6! shelving arrangements. First, arrange the three hardback books. There are 3! ways for this to occur. Once these are shelved, there are 4 places among the hardbacks to place the three pairs of paperbacks. These places can be chosen in C(4,3) ways. Finally, there are 6! ways to arrange the six paperback books.

II. In the expansion of (r + s + t + u + v)^15, determine the number of different ways a coefficient of 15 appears among the collected terms.

 Solution: We want to determine the number of times K=15 when we examine the collected terms of the form Kr^Rs^St^Tu^Uv^V where R+S+T+U+V=15 over the nonnegative integers. We know that K=(15!)/(R!S!T!U!V!), the multinomial coefficient. For K to be 15, it must be the case that R!S!T!U!V!=14! [Why is that?] For that to be true, two conditions must be met: (i) We must have exactly one of R, S, T, U, or V equal to 14 and (ii) the remaining 4 values must sum to 1. There are 5 ways for the first condition to be met: R or S or T or U or V must be 14. There are 4 ways for the second condition to be met: one of the four remaining values from among R,S,T,U,V must be 1 and the other three must be 0. By the multiplication principle, there are 5*4=20 total ways for a coefficient of 15 to appear.

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