Precalculus by Richard Wright

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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

10-07 Counting Principles

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.5.4

License Plate
Figure 1: License plate. credit (WallpaperFlare)

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.

Addition 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.

Multiplication Principle

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.

Distinguish Between Addition and Multiplication Principles

If the events are be combined with the word

Example 1: Counting Principles

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
  1. To choose 1 item you can choose 1 salad OR 1 entree. There is no overlap. This means addition principle.

    $$ 2 + 4 = 6 \text{ items} $$

  2. You are choosing 1 salad AND 1 entree. You choose the second event after the first event occurs. This means multiplication principle.

    $$ 2 × 4 = 8 \text{ meals} $$

Try It 1

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

Example 2: A Lock

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
  1. 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.

    $$ 100 · 100 · 100 = 1,000,000 $$

  2. 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.

    $$ 100 · 99 · 98 = 970,200 $$

Try It 2

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

Example 3: License Plates

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
  1. 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).

    $$ 26 · 26 · 26 · 10 · 10 · 10 = 17,576,000 \text{ license plates} $$

  2. Without repetition after the one letter or digit is chosen, then the next space has one less choice.

    $$ 26 · 25 · 24 · 10 · 9 · 8 = 11,232,000 \text{ license plates} $$

Try It 3

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

Permutations

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.

Permutation

The number of ways to order r objects out of n total different objects.

$$ _nP_r = \frac{n!}{(n-r)!} $$

Permutations on TI-84
  1. Enter the value of n
  2. Press MATH
  3. Use the arrow pad to move to the PRB menu
  4. Go down to select the nPr
  5. Enter the value of r
  6. Press ENTER

Entered as

n nPr r

Permutations on NumWorks
  1. Select Calculation from the home screen
  2. Press the Toolbox button
  3. Go down and select the Probability menu
  4. Select Combinatorics
  5. Select permute(,)
  6. Enter the value of n on left and the value of r on the right inside the parentheses
  7. Press EXE

Entered as

permute(n,r)

Example 4: Permutation

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

6 nPr 6 ENTER

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

Try It 4

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

Example 5: Permutation

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.

$$ _nP_r = \frac{n!}{(n-r)!} $$

$$ _{10}P_4 = \frac{10!}{(10-4)!} $$

$$ _{10}P_4 = \frac{10!}{6!} $$

$$ = \require{cancel} \frac{10·9·8·7·\cancel{6}·\cancel{5}·\cancel{4}·\cancel{3}·\cancel{2}·\cancel{1}}{\cancel{6}·\cancel{5}·\cancel{4}·\cancel{3}·\cancel{2}·\cancel{1}} $$

$$ = 10·9·8·7 = 5040 \text{ orders} $$

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

10 nPr 4 ENTER

On a NumWorks, enter

permute(10,4)

The output is 5040.

Try It 5

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

\(1.090273504×10^{14}\) 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 \(q_1, q_2, q_3, …\) 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.

Distinguishable Permutations (or Permutations with Repetition)

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 \(q_1, q_2, q_3, …\) are the number of times each object is repeated.

Example 6: Distinguishable Permutations

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 \(q_1 = 3\). T is repeated 2 times, so \(q_2 = 2\). No other letters are repeated. There are 6 total letters so \(n = 6\).

$$ \frac{n!}{q_1! q_2! q_3!…} $$

$$ \frac{6!}{3!2!} = 60 \text{ ways}$$

Try It 6

How many distinguishable ways are there to order the letters in the word HANNAH?

Answer

90 ways

Combinations

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.

Combination

The number of ways to group r objects out of n total different objects.

$$ _nC_r = \frac{n!}{(n-r)!r!} $$

Combinations on TI-84
  1. Enter the value of n
  2. Press MATH
  3. Use the arrow pad to move to the PRB menu
  4. Go down to select the nCr
  5. Enter the value of r
  6. Press ENTER

Entered as

n nCr r

Combinations on NumWorks
  1. Select Calculation from the home screen
  2. Press the Toolbox button
  3. Go down and select the Probability menu
  4. Select Combinatorics
  5. Select \(\left(\begin{matrix}n \\ k\end{matrix}\right)\)
  6. Enter the value of n on top and the value of r on the bottom
  7. Press EXE

Entered as

\(\left(\begin{matrix}n \\ r\end{matrix}\right)\)

Example 7: Combination

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 \(r = 12\).

$$ _{50}C_{12} = 1.213996511×10^{11} \text{ juries}$$

Try It 7

A 200 member church needs to select 15 deacons to help with church services. How many different groups of deacons can be selected?

Solution

\(1.462941635×10^{22}\) groups of deacons

Example 8: Combination

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

$$ _{15}C_4 · _{12}C_3 $$

$$ 1365 · 220 = 300,300 \text{ different dioramas} $$

Try It 8

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

\(1.52694773×10^{10}\) committees

Lesson Summary

Addition 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.


Multiplication Principle

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.


Distinguish Between Addition and Multiplication Principles

If the events are be combined with the word


Permutation

The number of ways to order r objects out of n total different objects.

$$ _nP_r = \frac{n!}{(n-r)!} $$


Permutations on TI-84
  1. Enter the value of n
  2. Press MATH
  3. Use the arrow pad to move to the PRB menu
  4. Go down to select the nPr
  5. Enter the value of r
  6. Press ENTER

Entered as

n nPr r


Permutations on NumWorks
  1. Select Calculation from the home screen
  2. Press the Toolbox button
  3. Go down and select the Probability menu
  4. Select Combinatorics
  5. Select permute(,)
  6. Enter the value of n on left and the value of r on the right inside the parentheses
  7. Press EXE

Entered as

permute(n,r)


Distinguishable Permutations (or Permutations with Repetition)

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 \(q_1, q_2, q_3, …\) are the number of times each object is repeated.


Combination

The number of ways to group r objects out of n total different objects.

$$ _nC_r = \frac{n!}{(n-r)!r!} $$


Combinations on TI-84
  1. Enter the value of n
  2. Press MATH
  3. Use the arrow pad to move to the PRB menu
  4. Go down to select the nCr
  5. Enter the value of r
  6. Press ENTER

Entered as

n nCr r


Combinations on NumWorks
  1. Select Calculation from the home screen
  2. Press the Toolbox button
  3. Go down and select the Probability menu
  4. Select Combinatorics
  5. Select \(\left(\begin{matrix}n \\ k\end{matrix}\right)\)
  6. Enter the value of n on top and the value of r on the bottom
  7. Press EXE

Entered as

\(\left(\begin{matrix}n \\ r\end{matrix}\right)\)

Helpful videos about this lesson.

Practice Exercises

    Decide whether the addition principle or multiplication principle should be used, then solve the problem.

  1. An ice cream shop offers a sundae with your choice of 1 of 5 flavors of ice cream and 1 of 3 flavors of syrup. How many different sundaes can be made?
  2. An ice cream shop offers 10 flavors of ice creams and 3 flavors of sherbet. A customer can buy a single scoop of ice cream or sherbet. How many choices does the customer have?
  3. An ice cream shop offers floats made with 1 flavor of ice cream and 1 flavor of pop. If they have 5 flavors of ice cream and 4 flavors of pop, how many different floats can they make?
  4. Answer the Counting Principle questions.

  5. A car manufacturer makes a model of car with 12 different colors of paint, 2 colors of interior fabric, and 3 exterior trim options. How many different looking cars can they produce?
  6. Your crazy teacher gives you a 5 question multiple choice quiz over tomorrow's lesson. Of course you don't know the answers, so you guess on all the questions. If each question has 4 choices, how many different sets of answers could you choose?
  7. How many different license plates can be made if each one is 1 letter followed by 3 digits if (a) with repetition and (b) without repetition?
  8. A combination lock opens with a correct entry of 3 numbers chosen from 60 numbers. How many different lock combinations can there be if (a) the numbers can be repeated and (b) the numbers cannot be repeated?
  9. Four couples bought tickets in the same row to a concert. How many different ways can they sit in the row (a) if it does not matter where they sit and (b) each couple sits together?
  10. Evaluate the permutation.

  11. \(_6P_4\)
  12. \(_3P_2\)
  13. Problem Solving

  14. A student club of 15 students is choosing its 5 officers. How many different arrangements of officers can there be?
  15. A group of 20 runners are randomly assigned lanes for their race. In the first race 6 runners are chosen. How many different orders can the runners in the six lane be selected?
  16. Three VeggieTales movies and 2 nature movies are going to be watched during the weekend. If the library has 9 VeggieTales and 10 nature movies to choose from, how many orders can the movies be watched?
  17. How many distinguishable ways can the letters in BOOKKEEPER be arranged?
  18. How many distinguishable ways can 3 identical starships, 2 identical planets, and 6 identical astronauts be arranged on a shelf?
  19. There are 13 players on a certain basketball team. The coach has decided that all the players will play every game. How many different groups of 5 players can the team be organized into (the positions are not important because the players can play all positions equally well)?
  20. Eight magazines are sitting on the end table. Freddi reads three of them. How many random groups of three magazines could Freddi have chosen?
  21. Students are to work in groups of four for a special project. If the class has 20 students, how many different groups of four could be created?
  22. A store received a shipment of 12 calculators contains 4 defective units. In how many ways can a school purchase 4 calculators and receive (a) 4 good units, (b) 3 good units, and (c) at least 2 good units.
  23. MegaMillions is a multistate lottery game. A player chooses 5 white balls from a set of 70 and 1 gold ball from a set of 25. The order that the balls are chosen is not important. (a) Find the number of possible winning MegaMillions numbers. (b) Find the number of possible winning MegaMillions numbers if they have to be chosen in order. (c) Compare this to a state lottery where the player chooses 6 balls from a set of 70.
  24. Mixed Review

  25. (10-06) Evaluate \(\left(\begin{matrix} 9 \\ 4 \end{matrix}\right)\).
  26. (10-06) Expand \(\left(2x + y\right)^4\).
  27. (10-05) Prove \(3 + 5 + 7 + 9 + \cdots + (2n + 1) = n(n + 2)\).
  28. (10-04) Write the rule for the nth term: 3, –12, 48, –192, ….
  29. (10-03) Find the sum of \(\displaystyle \sum_{n=1}^{50} 4n - 7\).

Answers

  1. Multiplication principle; 15
  2. Addition principle; 13
  3. Multiplication principle; 20
  4. 72
  5. 1024
  6. 26000; 18720
  7. 216000; 205320
  8. 40320; 384
  9. 360
  10. 6
  11. \(_{15}P_5 = 360,360\)
  12. \(_{20}P_6 = 27,907,200\)
  13. \(_9P_3 · \ _{10}P_2 = 45,360\)
  14. \(\frac{10!}{2!2!3!} = 151,200\)
  15. \(\frac{11!}{3!2!6!} = 4620\)
  16. \(_{13}C_5 = 1287\)
  17. \(_8C_3 = 56\)
  18. \(_{20}C_4 = 4845\)
  19. \(_8C_4 = 70; _8C_3 · _4C_1 = 224; _8C_2 · _4C_2 + _8C_3 · _4C_1 + _8C_4 = 462\)
  20. \(_{70}C_5 · _{25}C_1 = 302,575,350; _{70}P_5 · _{25}P_1 = 3.63×10^{10}; \ _{70}C_6 = 131,115,985\) a lot fewer options
  21. 126
  22. \(16x^4 + 32x^3y + 24x^2y^2 + 8xy^3 + y^4\)
  23. Show work and final step is \(k^2 + 4k + 3\)
  24. \(a_n = 3(-4)^{n-1}\)
  25. \(\frac{50}{2}\left(-3 + 193\right) = 4750\)