Illinois State University Mathematics Department
MAT 312: Probability and Statistics for Middle School Teachers Spring 1999 
Possible Solutions to Problem Set #3 . 
A. Car maintenance is not my favorite activity. Even the mundane chore of changing the oil is a pain. I usually wait longer than recommended to change the oil in my car. The owner of a local service station wanted to know whether other drivers had such bad habits, so he conducted a survey of his records to determine the length of time (in months) between customer oil changes. He randomly sampled station records and found the following 15 times (in months) between oil changes:

B. The table here shows cigarette consumption per adult per year and the number of deaths per 100,000 people per year from coronary heart disease (CHD) for 21 developed countries.

C. Counting Problems 
Questions 1113: All automobile license plates in Minnesota have six characters. The characters can be letters of the alphabet (A, B, . . . , Z) or singledigit numbers (0, 1, . . . , 9). 11. If a license plate has four letters followed
by two digits, with repetition allowed, how many
different license plates are possible? There are 26 choices for each letter and 10
choices for each digit. By the mutliplication
principle, we have 26*26*26*26*10*10=(26^4)*(10^2)
different license plates. 12. If a license plate again has four letters
followed by two digits, this time with repetition
not allowed, how many different license plates are
possible? Now we have 26 choices for the first letter, 25
for the next, followed by 24 and 23 choices. There
are 10 choices for the first digit and 9 for the
second. This yields 26*25*24*23*10*9 different
license plates. 13. If a license plate has three letters
followed by three digits, with no repetitions
allowed and no use of either the letter O or the
digit 0, how many different license plates are
possible? We start now with 25 letters and 9 digits
available, and repetition is not allowed. This
gives us 25*24*23*9*8*7 possible plates. 
Questions 1415: From a class of 20 students, a committee of 3 is to be formed to organize a fund raiser. 14. If all 20 students are eligible for
committee membership, how many different committees
could be formed? Because we are not concerned about any specific
committee offfices being filled nor about the order
that committee members are chosen, this is a
selection problem and we use a combination to
express the number of possiblilities: C(20,3). 15. If there are 12 women and 8 men in the
class, and the committee requires at least one
member of each gender, how many different
committees could be formed? There are now two ways to form a committee: two
women and one man or one woman and two men. The
first can be done in C(12,2)*C(8,1) ways, and the
second in C(12,1)*C(8,2) ways. Because those two
ways have nothing in common (we can't have a
committee with two women and one men be the same
committee as one with two men and one women), we
add the two possibilities to get a total of
C(12,2)*C(8,1) + C(12,1)*C(8,2) ways to create the
desired committee. You should convince yourself that the following
answer is WRONG: C(12,1)*C(8,1)*C(18,1), where
INCORRECT REASONING is to choose one woman from 12,
choose one man from 8, and choose one more from the
remaining 18. 
Questions 1621: Use Set I: {d,g,o}, Set II: {c,e,h,i,k,n}, Set III: {a,r,s,t} 16. One letter is to be chosen from Set I or Set
II or Set III. How many possible choices exist? There are no duplications among the elements in
the three sets, so we add the number of elements in
each set: 3+6+413 choices. 17. A 3letter set is to be created consisting
of one letter from each of Sets I, II, and III. How
many such 3letter sets are possible? We choose one letter from each set, which can be
done in 3*6*4=72 ways. 18. How many 6letter arrangements are possible
using the letters in Set II? This is a permutation of the letters in Set II.
There are P(6,6)=6! such arrangements. 19. How many 2letter sets can be made from the
letters in Set III? This is a selection not an arrangement, so we
use combinations: C(4,2)=6 such sets that can be
made. 20. How many 5letter arrangements can be made
such that the first and last letters are from Set I
and the other three letters are from Set III? We choose and arrange 2 of the three letters
from Set I (done in P(3,2)ways) and we choose and
arrange 3 of 4 letters from Set III (done in P(4,3)
ways). This results in P(3,2)*P(4,3)=12 possible
5letter arrangements. 21. How many arrangements of the letters in Set
II can be made so that no vowels are adjacent to
each other? In Set II there are two vowels and four
consonants. We use the consonants as "fences" to
separate the vowels. We do this do first arranging the four
consonants. this can be done in P(4,4)=4! ways.
With the consonants arranged, there now are five
"spaces" created among those consonants, including
at the ends of the line up. We therefore have 5
spaces to choose from in placing the first vowel
and 4 spaces remaining into which we place the
second vowel. thus, the vowels can be placed in 5*4
ways. Together, there are 4!*5*4=480 arrangements. 
Questions 2224: A postal worker entered Complex A of Reeseman Apartment Marketway and approached the row of mail boxes in the hallway. In a slot on each box was a label showing the name of the apartment dweller and a twodigit number for the apartment, as shown below.

Question 25: A fastfood chain advertises that it serves hamburgers in more than 1000 ways. The chain offers its burgers with various combinations of mustard, catsup, mayonnaise, Cajun spices, pickles, lettuce, cheese, and type of bun. 25. Is the chain's advertising claim legitimate?
Explain. In this discussion, I assume the following
possibilities, which may differ slightly from
yours. For each of the offerings described above, there
are a certain number of choices. most often, there
are two choices, either to have an item on the
sandwich or to not have it. I assume this is the
case for mustard, catsup,
mayonnaise, Cajun spices, pickles, and lettuce. So
for each of these 6 items there are two choices.
Together, then, there are 2*2*2*2*2*2=2^6=64 ways
to create a sandwich with or without these 6
items. There may be other options for cheeses and type
of bun. For instance, there may be three kinds of
cheese offered, and, of course, you can still have
the cheese withheld. For ther bun, I assume it is
not the case that an option is to NOT have a bun,
but I assume there are two or more choices for type
of bun. Now, to have more than 1000 ways you can get
your burger, we must have 64*c*b>1000, where c
represents the number of choices for cheese and b
represents the number of types of buns. Dividing
1000 by 64 shows that c*b must be 16 or larger. Therefore, the following possible pairs (c,b)
will work, as will an infinite number of
others: For instance, the first entry indicates there
are two choices for cheese (either have the one
kind they offer or have no cheese) and 8 types of
buns. So my explanation, given the assumptions
described here, shows it is possible and reasonable
that the claim is accurate of having more than 1000
ways to get a burger. 
Assigned: Tuesday 16 March 1999 Due: Tuesday 30 March 1999 