Illinois State University Mathematics Department


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

Summer 2006
Dr. Roger Day (day@math.ilstu.edu)


Test #2: Possible Solutions

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 J satisfies the equation ?

Solution: J = 7!

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

Solution: 10!/(4!2!2!2!)

(c) In the expansion of state:

  • (i) the number of uncollected terms; 
  • Solution: 3^8
  • (ii) the coefficient M in the collected term
  • Solution: M=8!/(2!5!1!)

(d) Determine the number of collected terms in the expansion of .

Solution: 41 collected terms

(e) Replace j and k in C(20,8) + C(20,9) = C(j,k) to correctly illustrate Pascal's Formula.

Solution: j = 21, k = 9 or C(21,9)

2.

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

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

Solution: 12!/(4!1!1!1!2!1!1!1!) = 12!/(4!2!)

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

Solution: (P(10,2)*10!)/(4!2!)

Arrange two of the 10 non-I letters in thefirst and last positions, P(10,2). Now permute the remaining 10 letters, 10!. Finally, divide by 4!2! to account for duplicates.

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

Solution: 6!*P(7,6)/(4!2!)

Place the 6 consonants first. There are 6! ways to do this. This placement creates 7 spaces within and outside the consonants. Permute the 6 vowels within these 7 available spaces, P(7,6), thereby assuring no two vowels are adjacent. Now divide by (4!2!) to account for duplication of the vowels.

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

Solution: C(12,6)

We need 12 places for the letters. Choose 6 of those places for the vowels. The remaining 6 places must contain the consonants.

3.

A ballot on a recent California election included 16 propositions. For each proposition, voters could choose AGREE or DISAGREE to express their preference.

(a) How many different marked ballots could be submitted, given that a choice of AGREE or DISAGREE had to be made for each of the 16 propositions? (3 points)

Solution: 2^16

For each of the 16 propositions, there are two choices to make. The decision for each proposition is independent of the other decisions.

(b) How many different marked ballots could show 9 propositions marked AGREE and 7 propositions marked DISAGREE, given that a choice of AGREE or DISAGREE had to be made for each of the 16 propositions? (3 points)

Solution: C(16,7) = C(16,9)

We need 16 proposition decisions. Choose 7 of those propositions and mark DISAGREE. The remaining 9 propositions must be marked AGREE.

(c) If voters could ABSTAIN from choosing a response on any of the propositions, how many different ballots could be submitted by voters? (4 points)

Solution: 3^16

For each of the 16 propositions, there are now three choices to make. The decision for each proposition is independent of the other decisions.

4.

On the floor are a pile of 9 mathematics books and 4 science books, all with different titles. The books are to be placed on one shelf.

(a) If a book's title distinguishes it from other books:

  • (i) How many ways can the 13 books be shelved? (2 points)
Solution: 13!
  • (ii) If no two science books may be adjacent, how many ways can the 13 books be shelved? (3 points)
Solution: 9!*P(10,4)

(b) If a book's subject (mathematics or science) is its only distinguishing property:

  • (i) How many ways can the 13 books be shelved? (2 points)
Solution: C(13,4) = C(13,9)
  • (ii) If all 4 science books must remain together, how many ways can the 13 books be shelved? (3 points)
Solution: C(10,1)

Place the 9 non-distinguishable mathematics books in a line on the shelf. this creates 10 spaces within and outside the mathematics books. Choose one of these 10 spaces and insert all 4 non-distinguishable science books.

5.

Consider the expansion of .

(a) How many uncollected terms are there? (3 points)

Solution: 4^25

(b) How many collected terms are there? (3 points)

Solution: C(28,3)

We need to determine how many terms are possible of the form m^Wa^Xt^Yh^Z, where W+X+Y+Z=25 and each of W,X,Y,Z is a non-negative integer. There are 25 total objects here with 4 types of objects, so the numbers of solutions to the equation is C(25+4-1,4-1).

(c) What is the coefficient of the collected term that contains the factor ? (4 points)

Solution: 25!/(10!4!6!5!)

6.

A domino is a rectangle formed by two congruent squares. Each square contains an orderly pattern of "pips" or dots representing a number from zero through nine. How many different dominoes can be made under these restrictions?

Solution: 55 dominoes

We consider two different cases that are disjoint.

I) The domino has a different number of pips in each of the two squares. In this case, there are 10 numbers possible for one square and 9 for the other. This yields 90 dominoes. Note, however, that the domino 5:8 is not distinguishable from the domino 8:5. Therefore, therefore half of the 90 dominoes counted here will be duplicates. We have 45 different dominoes each with a different pair of numbers.

II) The domino has the same number of pips in each of the two squares. There are ten such dominoes, from 0:0 to 9:9.

We have a total of 45+10=55 dominoes that can be created under the described conditions.


Syllabus
Grades & Grading
Content Notes
Session Outlines
Assignments and Problem Sets
Tests and Quizzes