Transcript PowerPoint Presentation - Unit 1 Module 1 Sets, elements

Part 1 Module 6 More counting problems
EXERCISE #1
From The FUNDAMENTALIZER, Part 2
www.math.fsu.edu/~wooland/count/count16.html
Mrs. Plato is going to order supper at a restaurant. She will choose items from each
of the following menu categories:
Category A: lettuce salad; cole slaw; clam chowder; potato salad; egg salad;
vegetable soup; bean soup; macaroni salad.
Category B: lamb; grouper fillet; pork chops; prime rib; shrimp; lobster; roast beef;
roast pork; fried chicken; baked chicken; broiled salmon.
Category C: baked potato; French fries; hash browns; rice; sweet potato; steamed
carrots; green beans; pilaf.
How many meal combinations are possible, assuming that Mrs. Plato will choose 3
items from Category A, 3 items from Category B, and 2 items from Category
C, and no item will be selected more than once?
A. 595,056
B. 18,627,840
C. 258,720
D. 43,614,208
Solution #1
This is a Fundamental Counting Principle problem, but it is more
complicated than those from Part 1 Module 6.
Here is a summary of Mrs. Plato’s three decisions.
Category A: Choose three distinct items from a list of eight items.
Category B: Choose three distinct items from a list of eleven items.
Category C: Choose two distinct items from a list of eight items.
Solution #1, page 2
This is a Fundamental Counting Principle problem, but it is more
complicated than those from Unit 3 Module 1.
Here are the number of options for each of the three compound decisions.
Category A: Choose three distinct items from a list of eight items.
C(8,3) = 56 different combinations
Category B: Choose three distinct items from a list of eleven items.
C(11,3) = 165 different combinations
Category C: Choose two distinct items from a list of eight items.
C(8,2) = 28 different options
According to the Fundamental Counting Principle, the number of outcomes is
56x165x28 = 258,720
Choice C is correct.
Exercise #2
From The FUNDAMENTALIZER, Part 2
www.math.fsu.edu/~wooland/count/count17.html
Mr. Moneybags, while out for a stroll, encounters a
group of ten children.
He has in his pocket four shiny new dimes and three
shiny new nickels that he will give to selected
children. In how many ways may these coins be
distributed among the children, assuming that no
child will get more than one coin?
A. 25,200
B. 3,628,800
C. 4,200
D. 604,800
Solution #2
Mr. Moneybags, while out for a stroll, encounters a group of ten
children.
He has in his pocket four shiny new dimes and three shiny new
nickels that he will give to selected children. In how many
ways may these coins be distributed among the children,
assuming that no child will get more than one coin?
This is a Fundamental Counting Principle problem, involving two
compound decisions.
Mr. Moneybags must:
1. Choose four children to receive the dimes, and
2. Choose three other children to receive the nickels.
Solution #2, page 2
Summary: He has ten children to choose from, and Mr.
Moneybags must:
1. Choose four children to receive the dimes
C(10,4) = 210 different combinations
2. Choose three other children to receive the nickels. These
means that he must choose three children from among the six
who didn’t get dimes.
C(6,3) = 20 different combinations
According to the Fundamental Counting Principle, the number of
outcomes is 210 x 20 = 4,200
The correct choice is C.
Exercise #3
There are nine waitresses and six busboys
employed at the trendy new restaurant The
House of Hummus. From among each of
these groups The International
Brother/Sisterhood of Table Service
Workers will select a shop steward and a
secretary.
How many outcomes are possible?
A. 51
B. 540
C. 2160
D. 102
E. None of these
Solution #3
This is a Fundamental Counting Principle problem
involving two compound decisions.
1. Choose a shop steward and a secretary from among
the nine waitresses.
2 . Choose a shop steward and a secretary from among
the six busboys.
Note that each of these two decisions involves
permutations, not combinations, because the
people being selected are getting different jobs or
titles.
Solution #3, page 2
1. Choose a shop steward and a secretary from among
the nine waitresses.
P(9,2) = 72 options
2 . Choose a shop steward and a secretary from among
the six busboys.
P(6,2) = 30 options
According to the Fundamental Counting Principle, the
number of outcomes is
72x30 = 2,160
Choice C is correct.
Why would we add?
In the examples we have looked at so far, we found
various numbers, perhaps by using the permutation
formula or the combination formula, and then
multiplied those numbers.
In some situations, however, we may want to add
numbers, rather than multiply, at the end of a
calculation.
The following example suggests why we might add,
rather the multiply.
Why would we add?
EXAMPLE
Suppose we ask, “How many of you are 18 years old?”
and, by show of hands, we see that there are 25
people who are 18 years of age.
Next, suppose we ask, “How many of you are 19 years
old” and, by show of hands, we see that there are
30 people who are 19 years of age.
Now we want to use those results to answer the
question “How many are 18 or 19 years old?”
It would make no sense to say that the number of
people who are 18 or 19 years old is 25x30 = 750.
Why would we add?
If 25 people are 18 years old, and 30 people are 19 years
old, then the number of people who are 18 or 19 years
old is
25+30 = 55.
This suggests the following fact:
To find the number of outcomes in an “either…or”
situation, find the number of of options for each case,
and add them.
More formally, if A, B are mutually exclusive
conditions, then
n(A or B) = n(A) + n(B).
“OR” means “ADD”
Generally, in counting problems or probability
problems,
“AND” means “MULTIPLY”
“OR” means “ADD”
Exercise #3a
There are nine waitresses and six busboys
employed at the trendy new restaurant The
House of Hummus. The International
Brother/Sisterhood of Table Service
Workers will either select a shop steward
and a secretary from among the waitresses,
or they will select a shop steward and a
secretary from among the busboys.
How many outcomes are possible?
A. 51
B. 540
C. 2160
D. 102
E. None of these
Solution #3a
This is an either/or scenario involving two mutually
exclusive cases
Either
Case 1. Choose a shop steward and a secretary from
among the nine waitresses.
or
Case 2 . Choose a shop steward and a secretary from
among the six busboys.
Note that each of these two cases involves
permutations, not combinations, because the
people being selected are getting different jobs or
titles.
Solution #3a, page 2
Case 1. Choose a shop steward and a secretary from
among the nine waitresses.
P(9,2) = 72 options
Case 2 . Choose a shop steward and a secretary from
among the six busboys.
P(6,2) = 30 options
Since we will perform either Case 1, or Case 2, but not
both, the number of outcomes is
72 + 30 = 102
Choice D is correct.
EXERCISE #4
A couple is expecting the birth of a baby. If the child is a girl, they will
choose her first name and middle name from this list of their
favorite girl’s names: Betty, Beverly, Bernice, Bonita, Barbie.
If the child is a boy, they will choose his first name and middle name
from this list of their favorite boy’s names: Biff, Buzz, Barney, Bart,
Buddy, Bert.
In either case, the child's first name will be different from the middle
name. How many two-part names are possible?
A. 50
B. 600
C. 61
D. 900
Solution #4
Girl’s names: Betty, Beverly, Bernice, Bonita, Barbie.
Boy’s names: Biff, Buzz, Barney, Bart, Buddy, Bert.
Any two-part name will be either a girl’s name or a boy’s name, so we
need to find the number of possible girl’s names, the number of
possible boy’s names, and then add these two numbers.
To form a two-part girl’s name, there are five options for the first name
and then four options for the middle name, so the number of
outcomes is 5x4=20 (this is the same as P(5,2)).
To form a two-part boy’s name, there are six options for the first name
and then five options for the middle name, so the number of
outcomes is 6x5=30 (this is the same as P(6,2)).
There are 20 possible two-part girl’s names, and 30 possible two-part
boy’s names, so the number of two-part names is 20+30=50.
Exercise #5
The mathematics department is going to hire a new instructor. They
want to hire somebody who possesses at least four of the
following traits:
1. Honest;
2. Trustworthy;
3. Loyal;
4. Gets along well with others;
5. Good at math;
6. Good handwriting
In how many ways is it possible to combine at least four of these
traits?
A. 360
B. 15
C. 22
D. 48
E. None of these
Solution #5
This is an either…or situation.
There are six traits to choose from. In this context, choosing at least
four traits means that we choose either four traits or five traits
or six traits.
Either:
1. Choose four traits: C(6,4) = 15
or
2. Choose five traits: C(6,5) = 6
or
3. Choose six traits: C(6,6) = 1
15+6+1 = 22 different ways to combine at least four of the six traits.
Exercise #6
Tonight there are 8 bus boys and 7 dishwashers
working at the trendy new restaurant Cap'n
Krusto's Crustacean Castle. Because it is a
slow night the manager will select either 3 bus
boys or 2 dishwashers and send them home
early.
How many outcomes are possible?
A. 14112
B. 378
C. 77
D. 1176
E. None of these
Solution #6
8 bus boys
7 dishwashers
Select either 3 bus boys or 2 dishwashers and send them home early.
In “either…or” situations we find the number of options for each case, and
then add.
The number of ways to choose 3 bus boys is C(8,3) = 56.
The number of ways to choose 2 dishwashers is C(7,2) = 21.
So, the number of ways to choose 3 bus boys OR 2 dishwashers is 56 + 21
= 77.
The correct choice is C.
Finally, if the problem had said that he was going to choose 3 bus boys and
2 dishwashers, then the answer would be 56x21 = 1176, choice D.
Exercise #7
The engineers at the Gomermatic Corporation have designed a new
model of automatic cat scrubbing machine, and a new model of
robotic dog poop scooper. Now it is the Marketing
Department's job to come up with catchy, high-tech sounding
names for the products. For each product they will randomly
generate a three-syllable name, such as Optexa, by choosing
one syllable from each of the following categories:
First syllable: Apt; Opt; Axt; Emt; Art; Ext.
Second syllable: a; y; e.
Third syllable: va; xa; ta; ra.
How many outcomes are possible, assuming that the two products
will not have the same name?
A. 5112
B. 143
C. 2556
D. none of these
Solution #7
First syllable: Apt; Opt; Axt; Emt; Art; Ext.
Second syllable: a; y; e.
Third syllable: va; xa; ta; ra.
There are two compound decisions. We must
1. Create a three syllable name for the automatic cat scrubbing machine; and
2. Create a three-syllable name for the robotic dog poop scooper.
Choosing a three-syllable name for the automatic cat scrubbing machine
requires three decisions:
Choose the first syllable: 6 options
Choose the second syllable: 3 options
Choose the third syllable: 4 options.
According to the Fundamental Counting Principle, the number of possible
names for the automatic cat scrubbing machine is (6)(3)(4) = 72. Having
chosen the name for the automatic cat scrubbing machine, there will then
be 71 possible names for the robotic dog poop scooper.
Thus, the total number of ways to name the two products is (72)(71) = 5112.
Exercise #8
Among a certain group of 28 mules, 23 are stubborn, 19
are obstinate, and 16 are both stubborn and obstinate.
How many are stubborn or obstinate?
A. 26
B. 51
C. 10
D. None of these
Solution #8
Among a certain group of 28 mules, 23 are stubborn, 19 are obstinate, and 16
are both stubborn and obstinate.
How many are stubborn or obstinate?
If a counting problem refers to conditions that are not mutually exclusive
(typically, a population described in terms of overlapping categories),
then we just draw a Venn diagram.
St.
7
Obs.
16
3
2
The Venn diagram shows that the number who are stubborn
or obstinate is 7+16+3 = 26
Remember: “Or” is always inclusive.
Exercise 9
In the Psychology Department there are 8 faculty
members.
2 faculty members will be chosen to design a new course
offering and 3 other faculty members will be chosen to
hang out in the faculty parking lot to prevent students
from parking there. How many different outcomes are
possible?
A. 560
B. 48
C. 240
D. None of these.
SOLUTION #9
This problem involves two dependent, compound decisions, among other
things.
First, choose two people to design a new course. There are eight people to
choose from; since the two people being chosen being assigned to the
same task, order doesn’t matter.
The number of ways to choose two people to design a new course offering is
C(8,2) = 28.
Next, having chosen two people to design a new course offering there will
remain six people to choose from for the job of hanging out in the faculty
parking lot to prevent students from parking there, so the number of ways
to choose those three people is C(6,3) = 20.
Thus, the number of ways to choose two people to design a new course
offering AND three other people to hang out in the faculty parking lot to
prevent students from parking there is
28x20 = 560. The correct choice is A.
New problem
In the Psychology Department there are 8 faculty members.
They will either choose 2 faculty members to design a new course offering or choose 3
faculty members to hang out in the faculty parking lot to prevent students from
parking there. How many different outcomes are possible?
Solution
For this either/or situation, we find the number of outcomes for each of the two cases,
and add.
If they decide to choose two people to design a new course, the number of outcomes is
C(8,2) = 28.
On the other hand, if instead they decide to choose three people to hang out in the
parking lot, they still have eight people from whom to choose, so the number of
outcomes is C(8,3) = 56.
Thus, the number of ways to either choose two people to design a new course offering
OR choose three people to hang out in the faculty parking lot is
28+56 = 84.