Precalculus by Richard Wright
Are you not my student and
has this helped you?
This book is available
to download as an epub.
Precalculus by Richard Wright
Are you not my student and
has this helped you?
This book is available
to download as an epub.
Again, the kingdom of heaven is like a merchant looking for fine pearls. When he found one of great value, he went away and sold everything he had and bought it. Matthew 13:45-46 NIV
Summary: In this section, you will:
SDA NAD Content Standards (2018): PC.5.4
There are millions of cars on the road and they all have to be licensed. Most places require a car to have a license plate with a unique arrangement of letter and numbers. The question is, with an increasing number of cars on the road each year, how many different license plates are there?
There are two basic counting principles. The addition principle and multiplication principle.
If one event can occur in m ways and a second event with no common outcomes can occur in n ways, then the first or second event can occur in m + n ways.
If one event can occur in m ways and a second event can occur in n ways after the first event has occurred, then the two events can occur in m × n ways. This is also known as the Fundamental Counting Principle.
If the events are be combined with the word
A restaurant offers 2 salads and 4 entrees. (a) How many different items can you choose if you get to choose 1 item? (b) How many different meals if you can choose 1 salad and 1 entree?
Solution
To choose 1 item you can choose 1 salad OR 1 entree. There is no overlap. This means addition principle.
2 + 4 = 6 items
You are choosing 1 salad AND 1 entree. You choose the second event after the first event occurs. This means multiplication principle.
2 × 4 = 8 meals
In a new job, Sally's employer offers three different laptops and five different tablets. (a) How many different items can Sally choose if she can only choose a laptop or tablet? (b) How many different choices does Sally have if she can choose 1 laptop and 1 tablet?
Answers
8 items; 15 choices
A lock will open with the correct choice of 3 numbers from 0 to 99 inclusive. How many different sets of 3 numbers are there if (a) the numbers can repeat or (b) the numbers cannot repeat?
Solution
All three numbers are needed. You need the first number AND the second number AND the third number. This is the multiplication principle. There are 100 choices for each number.
100 · 100 · 100 = 1,000,000
Without repeating means that after the first number is chosen, there is one less number available to choose. After the first two numbers are chosen, there are two less numbers to choose from.
100 · 99 · 98 = 970,200
A lock will open with the correct choice of 2 numbers from 0 to 49 inclusive. How many different sets of 2 numbers are there if (a) the numbers can repeat or (b) the numbers cannot repeat?
Answers
2500; 2450
How many different license plates are possible is each one is 3 letters followed by 3 digits (a) if the letters and digits can be repeated and (b) if the the letters and digits cannot be repeated?
Solution
The license plates numbers are chosen one after the other, so it is the multiplication principle. They are in the form L, L, L, D, D, D. There are 26 letters (A-Z) and 10 digits (0-9).
26 · 26 · 26 · 10 · 10 · 10 = 17,576,000 license plates
Without repetition after the one letter or digit is chosen, then the next space has one less choice.
26 · 25 · 24 · 10 · 9 · 8 = 11,232,000 license plates
How many different license plates are possible if each one is 2 digits followed by 4 letters (a) if the letters and digits can be repeated and (b) if the the letters and numbers cannot be repeated?
Answers
45,697,600; 32,292,000
Often there is a need to arrange a limited number of items. Even the license plate example without repetition is an arrangement problem. It is arranging 3 out of 26 letters and 3 out of 10 digits. Every time an object is chosen there is one less object to choose. For example, arranging 3 out of 26 letters would be 26·25·24 = 17,576 possibilities. This can be shortened using a permutation. A permutation gives the number of ways to order r objects out of n total objects where each object is different.
$$ _nP_r = \frac{n!}{(n-r)!} $$
There is a function on most calculators for permutations.
The number of ways to order r objects out of n total different objects.
$$ _nP_r = \frac{n!}{(n-r)!} $$
Entered as
n nPr r
Entered as
permute(n,r)
Six dogs are being lined up for the finals in a dog show. How many different orders can the six dogs be lined up?
Solution
This question is about ordering a limited number of different objects. Thus, a permutation will work. There are 6 dogs to choose from, so n = 6. All 6 dogs are being chosen, so r = 6. Calculate the permutation.
$$ _nP_r = \frac{n!}{(n-r)!} $$
$$ _6P_6 = \frac{6!}{(6-6)!} = 720 \text{ orders} $$
You might rather use the calculator function.
On a TI-84, type
¸ nPr ¸ Í
On a NumWorks, enter
permute(6,6)
The output is 720.
Notice that this gives the same output as using the multiplication principle. 6·5·4·3·2·1 = 720
Ten movies are being lined up on a shelf. How many different orders can the ten movies be lined up?
Answer
3,628,800 orders
Ten dogs have been entered into a dog show. How many different orders of 1st through 4th place can the dogs finish?
Solution
This is about ordering a limited number of objects, so it is a permutation problem. There are 10 total dogs, so n = 10. Four dogs are being chosen, so r = 4. Calculate the permutation.
$$ _nP_r = \frac{n!}{(n-r)!} $$
$$ _{10}P_4 = \frac{10!}{(10-4)!} $$
$$ _{10}P_4 = \frac{10!}{6!} $$
$$ = \require{cancel} \frac{10·9·8·7·\cancel{6}·\cancel{5}·\cancel{4}·\cancel{3}·\cancel{2}·\cancel{1}}{\cancel{6}·\cancel{5}·\cancel{4}·\cancel{3}·\cancel{2}·\cancel{1}} $$
= 10 · 9 · 8 · 7 = 5040 orders
This shows that a permutation uses the multiplication principle.
Of course, you could simply use your calculator to find the permutation.
On a TI-84, type
À Ê nPr ¶ Í
On a NumWorks, enter
permute(10,4)
The output is 5040.
There are 30 students who drive to school, but only 10 parking spaces close to the school. How many different orders can 10 of the 30 cars be arranged in the parking spaces?
Answer
1.090273504×1014 orders
If some of the objects of a permutation are repeated, the number of distinguishable permutations is
$$ \frac{n!}{q_1! q_2! q_3!…} $$
where q1, q2, q3, … are the number of times each object is repeated.
A distinguishable permutation is an arrangement that looks different from another arrangement. DAD and DAD look the same even though the D's have traded places. They are not distinguishable. DAD and ADD are distinguishable because they look different.
If some of the objects of a permutation are repeated, the number of distinguishable permutations of all the objects is
$$ \frac{n!}{q_1! q_2! q_3!…} $$
where q1, q2, q3, … are the number of times each object is repeated.
How many distinguishable ways are there to order the letters in the word TEETER?
Solution
There is the key word of "distinguishable" and some of the letters are repeated. E is repeated 3 times, so q1 = 3. T is repeated 2 times, so q2 = 2. No other letters are repeated. There are 6 total letters so n = 6.
$$ \frac{n!}{q_1! q_2! q_3!…} $$
$$ \frac{6!}{3!2!} = 60 \text{ ways}$$
How many distinguishable ways are there to order the letters in the word HANNAH?
Answer
90 ways
Sometimes there is a need to group a limited number of items. A combination gives the number of ways to group r objects out of n total objects without order where each object is different. Combinations group objects, but does not put the objects in order. This could be like choosing fish for an aquarium. The aquarium is the same no matter which order the fish are added. Combinations are the same as the coefficients for the binomial theorem from lesson 10-06.
$$ _nC_r = \frac{n!}{(n-r)!r!} $$
There is a function on most calculators for combinations.
The number of ways to group r objects out of n total different objects.
$$ _nC_r = \frac{n!}{(n-r)!r!} $$
Or another notation.
$$ \left(\begin{matrix} n \\ r \end{matrix}\right) = \frac{n!}{(n-r)!r!} $$
Entered as
n nCr r
Entered as
\(\left(\begin{matrix}\text{n} \\ \text{r}\end{matrix}\right)\)
A 12 member jury is selected out of a pool of 50 possible jurors. How many different juries can be selected (ignore that there is a foreman of the jury)?
Solution
A jury is a group of people where the order does not matter, so this is a combination problem. There are 50 jurors to choose form, so n = 50. Each jury is 12 people, so r = 12.
50C12 = 1.213996511×1011 juries
A 200 member church needs to select 15 deacons to help with church services. How many different groups of deacons can be selected?
Solution
1.462941635×1022 groups of deacons
Mr. Wright is making a diorama out of starships. He has 15 different Federation starships and 12 different Klingon starships. The diorama is to have any 4 Federation starships and any 3 Klingon starships. How many different dioramas can be created?
Solution
Order is not important because the problem does not indicate putting the ships in order. The problem indicates grouping the ships, so it is a combination. But, there are two groups; Federation and Klingon. The Klingon group will be chosen after choosing the Federation group. That indicates the multiplication principle. Thus, the combination of Federation starships is multiplied with the combination of the Klingon starships.
Federation Combination · Klingon Combination
15C4 · 12C3
1365 · 220 = 300,300 different dioramas
The school is forming a committee that consists of 4 seniors and 3 juniors. If there are a total of 58 seniors and 61 juniors, how many different committees could be formed?
Answer
1.52694773×1010 committees
If one event can occur in m ways and a second event with no common outcomes can occur in n ways, then the first or second event can occur in m + n ways.
If one event can occur in m ways and a second event can occur in n ways after the first event has occurred, then the two events can occur in m × n ways. This is also known as the Fundamental Counting Principle.
If the events are be combined with the word
The number of ways to order r objects out of n total different objects.
$$ _nP_r = \frac{n!}{(n-r)!} $$
Entered as
n nPr r
Entered as
permute(n,r)
If some of the objects of a permutation are repeated, the number of distinguishable permutations of all the objects is
$$ \frac{n!}{q_1! q_2! q_3!…} $$
where q1, q2, q3, … are the number of times each object is repeated.
The number of ways to group r objects out of n total different objects.
$$ _nC_r = \frac{n!}{(n-r)!r!} $$
Entered as
n nCr r
Entered as
\(\left(\begin{matrix}\text{n} \\ \text{r}\end{matrix}\right)\)
Helpful videos about this lesson.