Copyright © 2016 Looking Glass Ventures. All rights reserved. This book or parts thereof may not be reproduced in any form, stored in any retrieval system, or transmitted in any form by any means—electronic, mechanical, photocopy, recording, or otherwise— without prior written permission of the publisher, except as provided by United States of America copyright law. For permissions and additional information, write to the publisher, at
[email protected].
ISBN: 978-1-944931-02-5
Note To Reader Welcome to this little Zen Master’s guide on Probability, the second problem-solving book in our Zen series for middle-school students. As with all our texts in this series, our goal is to simply unveil the joys and delights of this mathematical topic, to provide context and make sense of the details, and help set you on a path of mathematical mastery and clever problem-solving. This title is of help and interest to students and educators alike. As with all the titles in the Zen Master’s series, this Probability guide is an eBook with a matching online course at http://edfinity.com/ZenSeries/Probability
(effective June 1, 2016).
So feel free to browse through this guide casually as a book on a digital device, or to work through its details as a focused, auto-graded course -- or both! All the details one needs to know about Probability are here in this book, explained naturally and swiftly, along with a robust compilation of practice problems. The more you try solving problems, the more confident you’ll become at them -- you’ll even start to notice recurring ideas and approaches which you can then use to your advantage. If you haven’t already, be sure to read 8 Tips to Conquer Any Problem in our Zen Master’s series. This is the must-read introduction to building your problem-solving skills. In fact, this guide is going to assume you
are familiar with the strategies and advice we go through there. Okay! Without further ado, let’s get started!
James Tanton March 2016
Acknowledgements My deepest thanks and appreciation to Michael Pearson, Executive Director of the Mathematical Association of America, for setting me on the path of joyous mathematical problem solving with the MAA Curriculum Inspirations project, and to Shivram Venkat at Edfinity for inviting me to extend that wonderful work to the global community of younger budding mathematicians. I am so very honored to be part of the unique, and truly remarkable, digital format experience Shivram and Edfinity have developed for the world.
James Tanton January 2016
Edfinity’s Zen Master’s Series Edfinity’s Zen Master’s series is a collection of 11 digital titles (6 for Middle School and 5 for High School) created for the modern educator and student. The titles are available only in digital form and consist of carefully crafted problem collections designed to help students master problem solving. Each title guides students through the themes of a specific topic (such as Algebra or Probability), presenting concise expository content, select examples illustrating specific problem solving techniques, and between 150200 problems expertly arranged to help the user achieve complete mastery. The volumes are each accompanied with optional access to an Edfinity ‘digital companion’ presenting all the
problems in the title as a self-paced, online course with auto-grading and performance analysis. Educators may enroll their students to track their progress, or students/parents may enroll individually. Access to the guides provides educators access to rich, supplemental problem collections for classroom use. The Zen Master’s Series is designed to serve broad usage by educators and students alike, offering substantive general enrichment, development of foundational skills in problem solving, and contest preparation. In addition to helping students prepare effectively for local and major international contests, the problems provide robust attention to standards and guidelines of the Common Core State Standards in
Mathematics (USA), GCSE (UK), Singapore’s Math curriculum, Australian Curriculum, and most other international syllabi.
ZEN MASTER’S MIDDLE SCHOOL SERIES 8 Tips to Solve Any Problem, by James Tanton Counting and Probability, by James Tanton Numbers and the Number System, by James Tanton Structure, Patterns and Logic, by James Tanton Relationships and Equations, by James Tanton Geometry, by James Tanton Solutions Manual for each title by James Tanton
ZEN MASTER’S HIGH SCHOOL SERIES Algebra, by David Wells Geometry, by David Wells Number Theory, by David Wells Discrete Mathematics, by David Wells Advanced Topics, by David Wells Solutions Manual for each title by David Wells Enroll at http://edfinity.com/ZenSeries/Probability (effective June 1, 2016) for online practice with scoring and complete solutions.
1. Counting Set Sizes BASIC JARGON A set is a collection of things. These things can be real things: • The set of all stuffed animals in my bedroom right now. • The set of all people on the world with exactly 257,340 hairs on their head. Or they could be abstract things that exist in our minds: • The set of all even numbers. • The set of all words that rhyme with “house.” People usually describe a particular set just in words, like we have done in the four examples above, but sometimes they like to list the elements of the set explicitly if it is easy to do.
It has become the convention to use curly brackets, { and } ,when you do this. For example,
• {red,blue,orange} is the set of the three colors red, blue, and orange. • {A,B,C ,…,Y ,Z } is the set of capital letters in the English alphabet. • {2,4,6,8,....} is the set of even whole numbers. COMMENT: NOTICE HOW THE DOTS ARE USED TO MEAN “KEEP DOING THE OBVIOUS THING.”
The order in which one lists elements of a set is considered unimportant. For example, {red,blue,orange} and {blue,red,orange} and {orange,red,blue} , and so on, all represent the same set.
Question: Would you want to list the elements of the set {2,4,6,8,....} in a different order?
Also, it is assumed that one doesn’t repeat the elements in the set. For example, writing {red,red,red,blue,blue,orange,orange,orange, orange,orange}
would be considered strange and people wouldn’t know what to make of it. The size of a set is the number of things in it. This can sometimes be a bit tricky to think about.
• The set {red,blue,orange} has size 3. • The set of capital letters in the English alphabet has size 26 . • The set of all even whole numbers is infinite. • The set of all people with brown hair who can sit at the bottom of the ocean for an hour while holding their breath is zero.
Sometimes people don’t even know the size of some sets!
• Consider the set of all multiples of six that are both one more and one less than a prime number. (Twelve, for example, is one such multiple of six. So is eighteen.) No one on this planet currently knows the size of this set. (Most mathematicians suspect it is infinite, but they don’t know for sure.)
One could draw a diagram represent a set. People usually draw circles to represent a set, but a blob of any shape is fine. For example, this set shows the number of my friends who like math, 120 of them, and the number who don’t, just 3. (What is wrong with those people?)
Often people like to denote sets with symbols, usually, but not always, capital letters. For example, someone might say:
Let E denote the set of all even whole numbers. (So E = {2,4,6,8,...} .) or Call the set of people with two arms and the set of all people with three P2 arms P3 .
The symbol used to denote a particular set isn’t important (although it is nice to choose a symbol that seems helpful for remembering what the set is).
By the way … each thing in a set is called an element of the set. For example, 16 is an element of the set of even numbers, and I am an element of the set of people with brown hair. ✔ Problems 1,2 - solve on Edfinity.
TECHNICAL STUFF: UNIONS AND INTERSECTIONS
The language and notation here gets fancy. We’ll explain the language through an example. Let H be the set of all people with brown hair. Let E be the set of all people with brown eyes. The union of two sets is the set of all objects that belong to at least one of the sets. The symbol ∪ is used to denote union. H ∪ E is the set of all people who
either have brown hair, or have blue eyes, or both. The intersection of two sets is the set of all objects that belong to both sets. The symbol ∩ is used to denote intersection.
H ∩ E is the set of all people with
both brown hair and brown eyes. This picture represents the set of brown-haired people and the set of brown-eyed people.
Beatrice, who happens to have brown hair, but not brown eyes, is represented as a dot sitting inside the blob marked H but not inside the blob marked E.
Jin-Pyo, who happens to have both brown hair and brown eyes, is represented by a dot that lies inside both blobs. Arman has red hair and blue eyes and is represented by a dot outside both blobs. The union and the intersection of H and E is represented by those points that lie in the shaded regions shown.
Sets need not intersect: • The set of all even whole numbers and the set of all whole numbers
that end with 5 have no elements in common. • The set of all cows and the set of all planets in the solar system do not intersect. • The sets {A,B,C } and {D,E ,F } do not intersect. Sets that do not intersect might be depicted as follows:
Comment: It is possible for two sets P and Q to satisfy P ∩Q = Q . Sets arrange as shown below have this property.
An example of two sets like this could be: P = the set of all people who have ever opened a math book. Q = the set of people who have opened this book. In this picture we also have P ∪Q = P .
✔ Problem 3 - solve on Edfinity.
SIZES OF UNIONS AND INTERSECTIONS
In a class of 20 students, 10 have a pet cat, 8 have a pet dog, and 5 have both. Here’s a diagram that shows this information – and more!
Do you see that 10 students do indeed have a pet cat? Do you see that 8 have a pet dog? We also see that 3 students have a dog but not a cat, and that 5 students have a cat but not a dog. Also, a total of 13 students own at least one pet of the types we are considering.
Warning: 10 students Warning! have a cat, 8 have a dog. It is tempting to say then that 10+ 8 = 18 students own at least one pet, which is not 13 ! Do you see the trouble with this? In the sum 10+ 8 = 18 those people that own both a cat and a dog got counted twice: once as a cat owner and once as a dog owner. This means that the number 18 is too big (which we know it is!)
WARNING:
WATCH OUT FOR THE PEOPLE IN THE INTERSECTION OF TWO SETS. THEY CAN MAKE ADDING THE NUMBERS CORRECTLY A BIT TRICKY.
Here’s a more general situation. Consider two sets P and Q :
The number a is the number of elements that are in P but outside of Q , b is the number of elements that belong to both P and Q , and c is the number of elements that are in Q but outside of P . Some questions: • If b = 0 , what is happening? (Look at the diagram again.) This means that that there are no elements in the intersection of the two sets and the picture should be more like:
• If c = 0 , what is happening? In this case there are no elements in Q that don’t belong to P . The picture should be more like:
• If a = 0 , what is happening? In this case there are no elements in P that don’t belong to Q . The picture should be:
• What is the meaning of the number a + b ? This is the total number of element of P . It is the size of P . • What is the meaning of the number b + c ? This is the size of Q . • What is the meaning of the number a + b + c ? This is the size of P ∪Q , the union. Now things can get tricky!
EXAMPLE: A set P has 10 elements and a set Q has 8 elements. If their union has 12 elements, how many elements belong to both P and Q ? Answer: If we look at the general picture:
We are being told:
And
a + b = 10 b + c = 8 a + b + c = 12.
But algebra is too hard!
There are 12 elements in total, and P has 10 of them. It must be that c = 2 . Since Q has 8 elements and c = 2 , we must have b = 6 . There are 6 elements in the intersection. MATH STRATEGY: Some people like to notice the following relationship. Look again at the picture:
We have:
( ) size Q = b + c ( )
size P = a + b
So
( )
()
size P + size Q = a + b + b + c .
But a + b + c = size ( P ∪Q ) and there’s an extra b , which is size ( P ∩Q ) .
( )
()
(
)
(
size P + size Q = size P ∪Q + size P ∩Q
)
Tip Some people memorize this formula to answer questions about sizes of unions and intersections. I personally don’t! I’d rather look at the picture and then nut my way through the mathematics, just like we did in the previous examples.
SUBSETS One final piece: By a subset of a set A we mean another set all of whose elements, if there are any, already belong to A . For example, if A = {a,b,c,d,e} , then {a,c,d} and {e} are subsets. So are {a,b,c,d,e} , the whole set itself, and {} ,the set with nothing in it. (Read the definition again carefully!) These last two examples are a bit weird and are sometimes called improper subsets. All other subsets, with at least one element, and not containing everything, are called proper subsets. If B is a subset of A , we might write: B ⊂ A . The picture that goes with this situation is:
✔ Problems 4-10 -solve on Edfinity.
2. Basic Probability
There are five vowels in the English alphabet: A, E, I, O, and U. (Let’s not include Y for now.) We like to believe that if we choose a letter of the alphabet at random the chances of it being a vowel is: 5 . 26
(As a decimal, this is about 0.19. As a percentage, this is 19% .) The “ 5” in the numerator is the number of vowels and the “ 26 ” in the denominator is the total number of letters we are choosing from. In general … Suppose we have a set of possible outcomes we are considering.
For example, in picking a letter there are 26 possible outcomes: A,B,C ,..,Y ,Z } . { For example, in tossing a coin would consider a set of two possible outcomes: {heads,tails} . In rolling a die, we would consider a set of set of six possible outcomes, the possible rolls: {1,2,3,4,5,6} . But suppose we hoping to see a particular type of outcome from among the given options. For example, in picking a letter we might be hoping for a vowel: A,E ,I ,O,U } . { For example, in flipping a coin we might be hoping to see a head: head } . { In rolling a die, we might be hoping to see an even number: {2,4,6} .
Then we like to say that the probability of seeing the option we want is the fraction: sizeof the set of outcomes we want . p= sizeof the set of all possible outcomes
(Notice that the numerator is never larger than the denominator.) For example, the probability of picking a vowel from the set of letters of the alphabet is:
{ {
size A,E ,I ,O,U
} }
5 p= = . 26 size A,B,C ,...,Y ,Z
The probability of seeing a head when tossing a coin is:
{
size head
}
1 p= = . 2 size head,tail
{
}
The probability of rolling an even number with tossing a die is:
{
}
size 2,4,6
3 1 p= = = . size 1,2,3,4,5,6 6 2
{
}
Note Some things to note: 1. The probability of an event is always a number between 0 and 1. 2. If p = 0 , that is, if a probability is zero, then this must be because there are no outcomes among the options we want to see. In other words, we will have zero chance of seeing the outcome we desire. For example, what are the chances of seeing a “ 7 ” when rolling a die? Well,
there is nothing we want to see among the possible options {1,2,3,4,5,6} . So
size nothing 0 p= = = 0. size{1,2,3,4,5,6} 6
3. If p = 1, that is, if a probability is one or 100% , then this must be because every option we could possibly see is something we want to see. For example, what are the chances of seeing a number smaller than 33 when rolling a die? All options we could see are acceptable, so p =
size{1,2,3,4,5,6} = 1. size{1,2,3,4,5,6}
Our definition of probability applies to geometry as well. Here the word “size” means area. EXAMPLE: A bone is buried somewhere in a square 10 yard by 10 yard garden. A dog is leashed at a post in the center of the yard on a leash that is 3 yards long. What are the chances that the bone is in the region of the yard the dog can reach?
Answer: The circle has area 9π and the entire garden has area 100 . Thus the region of ground the dog can explore represents the proportion:
9π ≈ 28.3% 100
of the yard. It seems reasonable to say, then, that the dog has a 28.3% chance of finding the bone. In general, it seems right to say: If a point is chosen at random in a region A , then the probability that it lands in a sub-region B is given by: area(B) p= . area A
( )
Again, p is a number between 0 and 1.
COMMENT:
PEOPLE LIKE TO SAY THAT LINE SEGMENTS HAVE NO AREA. (DOES THIS SEEM REASONABLE?) IN WHICH CASE CONSIDER THE FOLLOWING QUESTION:
QUESTION: A point is chosen at random in a square of side length 1. What are the chances that it lies on one of the diagonals of the square?
Since we like to believe line segments have no area, then the area of the two diagonal lines is 0 . In which case, the probability we seek is:
area(diagonals) 0 p= = = 0. area(square) 1
Does it seem reasonable to you to say that there is a zero percent chance that we would choose a point that lies absolutely, exactly, for certain, on a diagonal? ✔ Problems 11-16 -solve on Edfinity.
OR AND NOT Many textbooks make students think about specific actions associated with the words “or” and “not.” My advice is not to worry about the “rules” they teach and just use your common sense! ✔ Problems 17-18 -solve on Edfinity.
The general rule for “or” is:
General Rules
The probability for either A or B happening is sizeof theunion A ∪ B . p= sizeof the set of alloutcomes
The general rule for “not” is: The probability of A not happening is sizeof the set of items not in A . p= sizeof the set of alloutcomes
EXAMPLE: When rolling a die, the chance of rolling a multiple of three is:
(
)
size{3,6} 2 p multipleof three = = . size{1,2,3,4,5,6} 6
The chance of not rolling a three is:
(
)
p not multipleof three = 4 = 6
size{1,2,4,5} size{1,2,3,4,5,6}
.
Since all the multiples of three and all the non-multiples of three constitute the
entire set of possible outcomes, these probabilities add to 1. People notice:
( ) (
)
p A + p not A = 1 .
A “rule” some people like to memorize is:
(
)
( )
p not A = 1− p A
Again, there is nothing to memorize if you are willing to use your common sense. ✔ Problems 19-20 -solve on Edfinity.
LOTS OF TRIES Problem 20 suggests an idea that feels intuitively right: If a certain outcome of an experiment has a probability p% of occurring, then if you perform the experiment many many times, then you’ll see the outcome about p% of the time. For example: If I roll a die a million times, I’ll likely 1 see a “ 5” about of the time. 6 If I toss a coin a 90,000 times, I’d likely see about 45,000 heads. If roll die 1000 times and the number 6 came up 933 times, I would
probably conclude that the die is loaded (and that the chanced of rolling a six are about 93.3% .) ✔ Problems 21,22 -solve on Edfinity.
3. Compound Events Let’s start this section in a strange way. Suppose 100 people walk down a garden path that leads to a fork. Those who turn left go to house A, those who turn right to house B.
Assume that there is a 50% chance that a person will turn one way over another.
In this set-up we’d expect, essentially, 50 people to end up at house A and 50 people at house B. The following diagram of one hundred dots (for 100 people) depicts this outcome:
The number “ 100 ” here doesn’t matter. The point is that if a square is used to denote the entire population of people walking down the path, then half the area of the square (half the people) end up with result A, and the second half of the square with result B.
This is a very simple example. Let’s practice some more complicates scenarios. EXAMPLE: Folk walk down the following system of paths. Use the square model to compute the fraction of people that end up at house A, at house B, and at house C. (Assume that each choice encountered at a fork in the path is equally likely.)
Answer: At the first fork, a third of the people turn left, a third go straight, and a third turn right.
Of those that turn left, half go to house A and half go to house B. All those who go straight, go to house B. Of those who turn right, a third go to A, a third to B, and a third to C.
We have:
We see now that the proportion of people that end up in house A is 1 6
given as half of a third, that’s , plus 1 9
a third of a third, that’s . This 1 1 5 . 6 9 18
proportion is + =
The proportion of people that end 1 1 1 11 . 6 3 9 18
up in house B is: + + =
The proportion of people that end up in house C is:
1 2 = . 9 18
(Question: Does it make sense that these three answers add to 1?) CHALLENGE: People walk down the following system of paths. Use the square model to compute the fraction of people that end up at each house. (Again assume that each choice encountered at a fork in the path is equally likely.)
(Just so you have it, the answers are
5 6 2 5 , , , and .) 18 18 18 18
Now let’s bring this back to probability theory: EXAMPLE: I roll a die and then toss a coin. What are the chances of getting an even number followed by a head? Answer: Think of this as a pathwalking problem with two houses labeled “WANT” and “DON’T WANT.” The forks in the road represent the options that can occur (each, with 50% chance of occurring):
This leads to the square model diagram:
We see that the desired outcome represents one quarter (half of a half) of the square:
(
)
1 p even AND head = . 4
EXAMPLE: I toss a quarter, then I toss a dime, and then I roll a die. What are the chances of receiving HEAD, HEAD, and “5 or 6”? Answer: Here’s the garden path:
This gives the square model:
We have:
(
)
2 1 p head AND head AND {5,6} = of 6 2 1 2 1 1 1 of of the square= × × = . 2 6 2 2 12
Notice that each of the fractions in the final answer is a fraction corresponding to the probability of an individual outcome in the series of tasks: 2 is the chance of rolling a five 6
or a six with a die,
1 is the chance of flipping a 2
head with a dime, and
1 is the chance of flipping a 2
head with a quarter.
We see we get these fractions of a fraction of a fraction of a square – the product of these fractions. In general:
General Rule
If one performs one task and hopes to get outcome A , and then performs a second, completely unrelated, task and hopes to get outcome B , then the probability of seeing A and then B is the product of the individual probabilities:
(
) ( ) ( )
p A and then B = p A × p B .
The garden path model shows this.
✔ Problems 23,24 -solve on Edfinity.
The following example shows the advantage of thinking of probability questions in terms of running many, many runs of the experiment. EXAMPLE: I roll a red die and a blue die. What are the chances that the sum of the two numbers I see add to a multiple of four? Answer: Here are all the ways I could see a multiple of four: Red = 1, Blue = 3: The chances of seeing this are 1 1 1 × = . 6 6 36
Red = 2, Blue = 2 or 6: The chances of seeing this are 1 2 2 × = . 6 6 36
Red = 3, Blue = 1 or 5: The chances of seeing this are
1 2 2 × = . 6 6 36
Red = 4, Blue = 4: The chances of seeing this are 1 1 1 × = . 6 6 36
Red = 5, Blue = 3: The chances of seeing this are 1 1 1 × = . 6 6 36
Red = 6, Blue = 2 or 6: The chances of seeing this are 1 2 2 × = . 6 6 36
Suppose we roll the dice a large number of times, say 3600 times. Then we’d expect to see the first option (Red =1, Blue = 3) about one
thirty-sixth of the time, that is, about 100 times, the second option about 200 times, the third about 200 times, the fourth about 100 times, the fifth about 100 times, and the sixth about 200 times. Thus of the 3600 rolls, 100+ 200+ 200+100+100+ 200 = 900 of the rolls have sum a multiple of four. Thus the probability of what we seek is: 900 1 = . 3600 4
“WITH REPLACEMENT” AND “WITHOUT REPLACEMENT” Suppose a bag contains seven red balls and three yellow balls. I am about to pull out a ball and look at its color, and then pull out a second ball and look at its color. I can ask: What are the chances I will see two yellow balls? There are two tasks here, for sure. Task 1: Pull out a ball and look at its color. As there are ten balls and three are yellow, the chances that the first ball I see will be yellow is:
(
)
3 p first ball yellow = . 10
Task 2: Pull out a second ball and look at its color. Some people might assume that we took the first ball out and left it out. Other might assume we put the first ball back in and then picked again. The question as it stands in unclear! Mathematicians use the phrase “without replacement” to mean: don’t put back in whatever it is you just took out. The phrase “with replacement” means: put whatever you take out back in before you start the next task. Here’s what I really meant to ask: EXAMPLE: A bag contains seven red balls and three yellow balls. I will pull two balls out of the bag, one after the other, with replacement. What are the chances I will see two yellow balls?
Answer: Task 1: Pull out a ball and look at its color.
(
)
3 p first ball yellow = . 10
Task 2: Pull out a ball and look at its color. Since we put the first ball back, there are again ten balls in the bag, three of which are yellow. And we have:
(
)
3 p second ball yellow = . 10
Thus the probability of seeing two yellow balls is:
(
)
3 3 9 p both yellow = × = . 10 10 100
Of course, now we wonder about the version of this question “without replacement.” A bag contains seven red balls and three yellow balls. I will pull two balls out of the bag, one after the other, without replacement. What are the chances I will see two yellow balls? Again, the probability of first seeing a yellow ball is p( first ball yellow ) =
3 . But 10
the probability of seeing a second yellow ball depends on what happened with the first ball! If the first ball seen really was yellow, then a yellow ball is taken away and there are two yellow balls left in the bag of nine balls. In this case: p( second ball yellow ) = .
2 9
If the first ball seen was not yellow, then a red ball is taken away and there are three yellow balls left in the bag of nine balls. In this case:
(
)
3 p second ball yellow = . 9
Which probability should we use? Let’s go back to the garden path model.
Imagine sending people down the path shown.
Which turn they make on the first fork depends on what color ball they first pick from the bag. We have that them will pick a yellow ball and continue on their journey. (The remaining
3 of 10
7 get a red ball and 10
immediately go to the DON’T WANT house.) Of those that are still on the garden path, having chosen a yellow ball first, their next turn is decided by the pick of a second ball, without replacement. All these people currently have a bag with nine balls in it, two of which are yellow. Thus each person has a
2 9
chance of going on to the WANT house.
So, all in all, two-ninths of the threetenths of the people will end up in the yellow-yellow WANT house. And what fraction of the square is this? 2 3 6 1 It is × = = of the square. 9 10 90 15
Thus, without replacement:
(
)
1 p both yellow = . 15
Advice
ADVICE:
“WITHOUT REPLACEMENT” PROBLEMS TEND TO BE TRICKY! IT REALLY DOES HELP TO DRAW A GARDEN PATH MODEL FOR THEM, KEEPING TRACK OF THE FRACTION OF PEOPLE WHO LUCKILY MAKE ALL THE CHOICES YOU ARE HOPING FOR..
Question: Can you also think about this second, “without replacement,” question by imagining performing the experiment of choosing two balls, say, 900 times? Can you see the answer
again?
1 15
(In general, is this approach of imaging performing a large number of runs of the experiment actually the same as garden-path approach?)
AVOID SIMULTANEOUS ACTIONS!
In tossing a coin and rolling a die, say, there are three possibilities: 1)Toss the coin & then roll the die, 2)Roll the die & then toss the coin, 3)Roll & toss the coin simultaneously. Standing back and thinking about this one would say that the coin and the die, once they’ve been flipped and tossed, don’t “know” or “care” in which order their outcomes happened: all three scenarios are philosophically equivalent. So we have some choice on how to think about how actions are performed.
Here’s my advice:
Advice
AVOID SIMULTANEOUS THINKING: ALWAYS IMAGINE THAT ONE ACTION OCCURS FIRST AND ANOTHER ONE SECOND. (AND IT USUALLY DOESN’T MATTER WHICH ONE YOU THINK OF AS OCCURING FIRST.)
This is important for examples like the following: EXAMPLE: Two dice are tossed. What are the chances of rolling two values that sum to 7 ? Answer: When people think of tossing two dice they think of doing so simultaneously. In which case it looks like there are three ways to roll a sum of 7 :
Roll a 1 and a 6 . Roll a 2 and a 5. Roll a 3 and a 4 . Or is it more than this? Rolling a 4 and a 3? Is that the same as rolling a 3 and a 4 ? Warning! Simultaneous thinking is confusing. Avoid it! The dice don’t “care” if they land on the ground one at a time or simultaneously. So let’s go with the “one at a time” idea. (Actually, in real life, nothing happens simultaneously: one die will always stop rolling ever so slightly before the second one stops. There is always a first die and a second die!)
It is clear with a first and a second roll that there are SIX ways to roll a sum of 7 . First roll = 1; Second roll = 6 . First roll = 2; Second roll = 5. First roll = 3; Second roll = 4 . First roll = 4 ; Second roll = 3. First roll = 5; Second roll = 2. First roll = 6 ; Second roll = 1. How many possible outcomes are there in total? The first roll can be any one of six numbers, as can be the second roll. There are thus a total of 6 × 6 = 36 results we could see in rolling a first and second die. Only the six options above give us a sum of 7 , so:
(
)
6 1 p rolling7 = = . 36 6
EXAMPLE: I roll a pair of dice. What are the chances of seeing a 6 and a 2? Answer: Assume one die is rolled first and the other second. There are a total of 36 outcomes we could see. Now we are hoping to see: or
First roll = 6 ; Second roll = 2, First roll = 2; Second roll = 6 .
The probability is thus p =
2 1 = . 36 18
✔ Problems 25-32 -solve on Edfinity.
4. The Multiplication Principle For Counting Let’s start with a tiny puzzle: Suppose that there are three major highways from Adelaide to Brisbane, and four major highways from Brisbane to Canberra.
How many different routes can one take to travel from Adelaide to Canberra via Brisbane (with no backtracking)?
Here’s the real question:
Is the answer to this puzzle 7 , coming from 3 + 4 , or is the answer 12 , coming from 3 × 4 ? We need to be very clear as to whether we add or multiply these numbers. Here’s one way to think about it: If we take the “top” route from A to B , then there are 4 options as to which highway to take next:
If we take the “middle” route from A to B , then there are 4 options as to which highway to take next:
And if we take the “bottom” route from A to B , then there are 4 options as to which highway to take next:
Thus there are a total of 4 + 4 + 4 routes from A to C . Notice that we have three groups of four and, in basic arithmetic, repeated addition is multiplication. So the answer is: three groups of four = 3× 4 = 12.
It is appropriate to multiply the numbers given. EXAMPLE: Suppose that there are six major highways from Canberra to Darwin:
How many different routes are there from A to D (with no backtracking)? Answer: For each of the 12 routes from A to C there are 6 options of which highway to take next from C to D . Twelve groups of six gives 12× 6 = 72 options.
This answer is actually the product: 3 × 4 × 6 . We see multiplication at play all the way through! EXAMPLE: a) I own five different shirts and four different pairs of trousers. How many different outfits could you see me in? b) Actually, I own five different shirts, four different pairs of trousers, and two sets of shoes. How many different outfits could you see me in? Answer: a) For each shirt I have 4 options for which pair of trousers to wear. Thus I have a total of 4 + 4 + 4 + 4 + 4 = 5× 4 = 20 different outfits available to choose from. b) I have 5 × 4 × 2 = 80 different outfits if I consider my shoes! Comment: This previous example assumes that there are no restrictions
on which shirts, trousers, and shoes I might wear. For example, if I will never wear my red shirt with my purple trousers and green shoes, then there are only 79 possible outfits you could see me in. We have a general principle:
THE MULTIPLICATION PRINCIPLE FOR COUNTING: IF THERE ARE TASK AND
WAYS TO COMPLETE ONE
WAYS TO COMPLETE A SECOND
TASK, AND THE OUTCOMES OF THE FIRST TASK IN NO WAY AFFECT THE CHOICES MADE FOR THE SECOND TASK, THEN THE NUMBER OF DIFFERENT WAYS TO COMPLETE BOTH TASKS IS
.
This principle readily extends to the completion of more than one task.
EXAMPLE: On a multiple choice quiz there are five questions, each with four choices for an answer:
I decide to fill out my answers randomly. In how many different ways could I fill out the quiz? Answer: This is a five-task process: Task 1: Answer question one: 4 ways Task 2: Answer question two: 4 ways Task 3: Answer question three: 4 ways
Task 4: Answer question four: 4 ways Task 5: Answer question five: 4 ways By the multiplication principle there are 4 × 4 × 4 × 4 × 4 = 45 = 1024 ways to complete the quiz. EXAMPLE: On a multiple choice quiz there are five questions, each with four choices for an answer:
I decide to fill out my answers random make a pattern. In how many different ways could I fill out the quiz so that the second answer is different from the first, the third answer is either A or D, and the last two answers are the same? Answer: Again this is a five-task process: Task 1: Answer question one in any way I like: 4 ways Task 2: Answer question two in any way different from answer one: 3 ways Task 3: Answer question three with either A or D: 2 ways Task 4: Answer question four in any way I like: 4 ways Task 5: Answer question five in the same way as answer four: 1 way By the multiplication principle there are 4 × 3× 2× 4 ×1 = 96 ways to complete the quiz this way. ✔ Problems 33-38 -solve on Edfinity.
5. Arranging Letters: Factorials The next three sections describe content that doesn’t often appear in middle-school mathematics competitions, at least not directly. (There are often questions related to these ideas though.) But this does not mean you can’t look at this material and use it to your advantage! It will make those related questions easier. Plus, this mathematics content is just cool and fun in and of itself! EXAMPLE: My name is JIM. In how many ways can I rearrange the letters of my name? Answer 1: We can just list out all the possibilities: JIM JMI
MJI MIJ
IJM IMJ
There are six ways.
Answer 2: Let’s use the multiplication principle: We have three slots to fill with letters:
This is really a set of three tasks. Task 1: Fill the first slot with a letter. There are 3 ways to complete this task. (Fill the slot with J, or with I, or with M.) Task 2: Fill the middle slot with a letter. There are 2 ways to complete this task. (Once the first slot is filled, there are only two choices of letters left to use for the second slot.)
Task 3: Fill the final slot with a letter. There is only 1 way to complete this task (once slots one and two are filled).
By the multiplication principle, there are thus 3 × 2×1 = 6 ways to complete this task. (Was the first approach easier?) EXAMPLE: Actually my proper name is JAMES. In how many ways can I rearrange the letters of my proper name? Answer 1: Listing out all the possibilities doesn’t seem fun any more. Let’s forget answer 1!
Answer 2: We have five slots to fill, a series of five tasks:
There are 5 ways to place a first letter, 4 ways to place a second letter, 3 ways for a third letter, 2 choices for the fourth letter, and 1 option for the final letter. This gives a total of 5 × 4 × 3× 2×1 = 120 ways to write out the letters of JAMES. Question: In how many ways can one write out the letters of BOVINE? (Do you see that the answer is 720?) Question: In how many ways can one arrange the letters of OF? (Do you see that the answer can be thought of as 2 ×1 ?)
In playing with these problems it seems the following types of products keep coming up: 2×1 = 2 3× 2×1 = 6 4 × 3× 2×1 = 24 5× 4 × 3× 2×1 = 120 6 ×5× 4 × 3× 2×1 = 720
Mathematicians give these products a name: DEFINITION: THE PRODUCT OF THE INTEGERS FROM TO (OR EQUIVALENTLY, FROM DOWN TO ) IS CALLED FACTORIAL AND IS DENOTED .
For example, 3! = 1× 2× 3 = 6 . (Or we could write: 3! = 3× 2×1 = 6 .)
and
5! = 1× 2× 3× 4 ×5 = 120 .
Mathematicians have decided to declare: 1! = 1 .
(Does it make sense to say that the product of all the numbers from 1 up to 1 is 1?) These factorial numbers grow very large very quickly: 1! = 1 2! = 2×1 = 2 3! = 3× 2×1 = 6 4! = 4 × 3× 2×1 = 24 5! = 5× 4 × 3× 2×1 = 120 6! = 6 ×5× 4 × 3× 2×1 = 720 7! = 7 × 6 ×5× 4 × 3× 2×1 = 5040 8! = 8 × 7 × 6 ×5× 4 × 3× 2×1 = 40320 9! = 9× 8 × 7 × 6 ×5× 4 × 3× 2×1 = 362880 10! = 10× 9× 8 × 7 × 6 ×5× 4 × 3× 2×1 = 3628800
Question: What is the first factorial larger than a billion? Question: Most calculators have a feature that let’s you compute factorials. What is the largest factorial your calculator can handle? ✔ Problems 39-42 -solve on Edfinity.
6. Dealing With Repeated Letters We have: THERE ARE N! WAYS TO ARRANGE A STRING OF N DISTINCT LETTERS.
My name, be it JIM or JAMES, contains no repeat letters. But what if my name were BOB, with the letter B repeated? EXAMPLE: How many ways are there to arrange the letters BOB if the B s are identical and cannot be distinguished? Answer: This name is small enough for us to just list all the possibilities: BOB
BBO
OBB
There are three ways to arrange BOB.
How about a bigger word with repeat letters? EXAMPLE: In how many ways can one arrange the letters HOUSES ? Listing all the possibilities seems overwhelming this time. Hmm. Strategy Here’s a clever approach: CONVERT THE PROBLEM TO ONE WE ALREADY KNOW HOW TO SOLVE!
We know how to count the arrangements of words if all the letters in the word are different. For example:
There are 5! = 120 ways to arrange the letters of HOUSE . We can solve the problem if one of the S s is missing. We can also solve the problem if the two S s were different, say, S1 and S2 . There are 6! = 720 ways to arrange the letters of HOUS1ES2 . Let me start listing all 720 ways with two different S s:
HOUS1ES2 HOUS2ES1 S1OUS2HE S2OUS1HE EOHUS1 S2 EOHUS2S1
!
But notice! If the S s are no longer distinguishable, then pairs in this list of answers “collapse” to give the same arrangement.
This shows we must alter our answer of 720 by a factor of two if we make the S s the same. 720 = 360 ways to arrange 2 the letters HOUSES .
There are
Question: How many ways are there to arrange the letters BABE ? (Check you answer by writing out all the possibilities!) Let’s go up a notch in difficulty: EXAMPLE: How many ways are there to rearrange the letters of the word CHEESE ? Answer: If the three E s were distinct – written E1 , E2 , and E3 , say – then there would be 6! = 720 ways to rearrange the letters CHE1E2SE3 . But the three E s can be rearranged 3! = 6 different ways within any one
particular arrangement of letters. These six arrangements would be seen as the same if the E s were no longer distinct:
Thus we must divide our answer of 6! by 3! to account for the groupings of six that become identical. There are thus
6! 6⋅5⋅4⋅3⋅2⋅1 = = 6⋅5⋅4 = 120 3⋅2⋅1 3!
ways to arrange the letters of CHEESE . Let’s make it even worse!
EXAMPLE: How many ways are there to arrange the letters CHEESES ? Answer: If the E s were distinguishable and the S s were distinguishable, then we’d be counting the ways to arrange seven distinct letters: There are 7! ways to arrange the letters of CHE1E2S1E3S2 As before there are 3! ways to arrange the E s in any particular configuration. These groups of 3! will “collapse” to the same arrangement if we remove the subscripts from the E s.
But these new arrangements also collapse in pairs once we remove the subscripts from the S s.
So we need to take our answer of 7! and divide it by 3! and again by 2! .
We get the funny looking fraction: 7! 3! 2!
Let’s multiply the top and bottom lines each by 3!. This now reads: 7! × 3! 7! 3! = . 2!× 3! 2!3!
which is much easier to read! And this number equals 7⋅6⋅5⋅4⋅3⋅2⋅1 7⋅6⋅5⋅4 = = 21⋅20 = 420 . 2 3⋅2⋅1⋅2⋅1
Now we’re pros!
Question: In how many ways can one arrange the letters CHEESIEST ? Answer:
9! , whatever that 3!2!
number is.
Consider the “word” CHEESIESTESSNESS , the quality of being the cheesiest of cheeses of all. Do you see that there are 16! ways to rearrange its letters? 5!6!
It is actually better to write this answer as
16! 1!1!5!6!1!1!1! 1! 1! 5! 6! 1! 1! 1!
for the one letter C . for the one letter H . for the five letters E . for the four letters S . for the one letter I . for the one letter T . for the one letter N .
This offers a self-check: The numbers appearing on the bottom sum to the number appearing on the top. (This is why mathematicians like to regard 1! as equal to 1. It makes doing this convenient and consistent.) One more example: EXAMPLE: In how many ways can we arrange the letters AAAABBCDDD ? Answer: There are 10 letters in total, and if we could tell all the letters apart (for A1 A2 A3 A4 B1B2CD1D2D3 ) the answer would be 10! . But in each of these arrangements, there are 4! arrangements of A1 , A2 , , and A4 that “collapse” to the same
arrangement when the A s are made to look the same. And there are 2! arrangements of B1 and B2 that “collapse” to the same arrangement when the B s are made to look the same. And 1! arrangements of the single C (this isn’t saying much), and 3! arrangements of D1 , D2 , and D3 that “collapse” to be the same when the D s are made to look the same. So we need to divide our answer of 10! by 4! and by 2! and by 1! and by 3! . There are 10! 4!2!1!3!
ways to arrange the letters AAAABBCDDD . (Notice that the
numbers in the denominator add to the number in the numerator? We have accounted for all the letters.) If we want the actual number (do we?), this is 10! 10× 9× 8 × 7 × 6 ×5× 4 × 3× 2×1 = 4!2!1!3! 4 × 3× 2×1× 2×1×1× 3× 2×1 10× 9× 8 × 7 × 6 ×5 = 2× 6 = 10× 9× 4 × 7 ×5 = 12,600
We now have the general principle: WORD ARRANGEMENT PRINCIPLE Suppose a set of N letters has a copies of one letter, b copies of a next letter, and so on, all the way up to z copies of a final letter, then there are N! a!b!!z!
ways to arrange those N letters.
(The numbers in the denominator sum to the number in the numerator.) 10! For example, there are 4!2!1!3! arrangements of AAAABBCDDD .
PROBLEM 43: How many arrangements of the letters DOODLENOODLE are there with the letter D appearing at the start and the letter O appearing at the end?
e
7. Letter Arrangements In Disguise So we know how to answer questions of the type: How many ways can you rearrange the letters of AAABBBBCCCCCC? 13! Answer: , whatever that 3!4!6!
number turns out to be.
Now look at this next problem. Can you see that it actually the same question in disguise? EXAMPLE: Mean Mr. Muckins has a class of 13 students. He has decided to call three of the students A students, four of them B students, and six of them C students before
the year has even started! In how many ways could he assign these labels? Answer: Let’s imagine all thirteen students are standing in a line: Here’s one way he can assign grades: Here’s another way: and so on. We see that this grade question is just the same problem as rearranging letters the letters AAABBBBCCCCCC. The answer must be Cool!
13! . 3!4!6!
EXAMPLE: There are 10 people in an office and 4 are needed for a committee. In how many different ways can we make a committee of four? Answer: Imagine the 10 people standing in a line. We need to give out labels. Four people will be called “ON” and six people will be called “LUCKY.” (You don’t really want to be on a committee. It is usually extra work!) Here is one way to assign those labels:
And we can list many more ways. But we see that this is just a letter arrangement problem. How many ways can we arrange the letters OOOOLLLLLL ?
The answer is: 10! 10⋅9⋅8⋅7 = = 10⋅3⋅7 = 210 . 4!6! 4⋅3⋅2⋅1
EXAMPLE: Fifteen horses run a race. How many possibilities are there for first, second, and third place? Answer: Imagine the horses lined up in a row. One horse will be labeled “first,” one will be labeled “second,” one “third,” and twelve will be labeled “losers.” Here’s one possibility for how the race might turn out.
And there are many more possibilities. But we see counting all possibilities is just answering the question:
How many ways can we arrange 123LLLLLLLLL? The answer is: 15! = 15×14 ×13 = 2730 . 1!1!1!12! COMMENT: IT IS IMPORTANT THAT EACH AND EVERY OBJECT GET A LABEL. WE HAVE TO CONSIDER THE HORSES THAT DON’T WIN AS WELL.
EXAMPLE: A “feel good” running race has 20 participants. Three will be deemed equal “first place winners,” five will be deemed “equal second place winners,” and the rest will be deemed “equal third place winners.” How many different outcomes can occur?
Answer: Do you see that the 20! ? 3!5!12!
answer is
Do you see that
the problem is the same as asking for how many ways to arrange a string of 20 letters? EXAMPLE: From an office of 20 people, two committees are needed. The first committee shall have 7 members, one of which shall be the chair and one the treasurer. The second committee shall have 8 members. This committee will have three co-chairs and two co-secretaries and one treasurer. In how many ways can this be done (assuming no one can be on two committees, nor have two or more roles within a committee)? Answer: First let’s keep track of the labels. 1 person will be labeled “chair of first committee,” C1 .
1 person will be labeled “treasure of first committee,” T1 . 5 people will be labeled “ordinary members of first committee,” O1 . 3 people will be labeled “co-chairs of second committee,” C2 . 2 people will be labeled “cosecretaries of second committee,” . S2 1 person will be labeled “treasurer of second committee,” T2 . 2 people will be labeled “ordinary members of the second committee,” . O2 5 people will be labeled “lucky,” they are on neither committee, L . If we line the people up in a row, then each arrangement would look like an arrangement of the letters: . C1T1O1O1O1O1O1C2C2C2S2S2T2O2O2LLLLL
The total number of possibilities is thus
20! , which we can 1!1!5!3!2!1!2!5!
work out as a number if we want. School-book writers will be shocked that I haven’t used the fancy words “permutation” and “combination” in these notes. We don’t need them! (If you have heard these words before, feel free to forget them. If you haven’t heard these words before, you are not missing out on anything!) EXAMPLE: (A “combination” problem) Suppose 5 people are to be chosen from 12 and the order in which folk are chosen is not important. How many ways can this be done? Answer: If we line people up in a row, five will get the label C for chosen, and seven the label N for
not chosen. Here’s one possible way to choose five from the twelve.
Each possibility is just an arrangement of the letters: CCCCCNNNNNNN. There are thus 12! ways to accomplish this task. 5!7!
(By the way: 12! 12×11×10× 9× 8 = = 12×11× 3× 4 = 1584 .) 5× 4 × 3× 2×1 5!7!
EXAMPLE: (A “permutation” problem) Suppose 5 people are to be chosen from 12 for a team and the order in which they are chosen is considered important. In how many ways can this be done? Answer: This time we have the labels: 1 person labeled “first” 1 person labeled “second”
1 person labeled “third” 1 person labeled “fourth” 1 person labeled “fifth” 7 people labeled “not chosen” If we line the people in a row, one way of choosing five people, in order, could be:
Each possibility is just an arrangement of the letters: 12345NNNNNNN . 12! And there are such 1!1!1!1!1!7!
arrangements.
This is the answer. (And the actual number is: 12! = 12×11×10× 9× 8 = 95040 .) 1!1!1!1!1!7!
Question: Does it make sense that the answer to the second problem is larger than the first? ✔ Problems 44-50 -solve on Edfinity.
About The Author
JAMES TANTON Visit http://www.maa.org/mathcompetitions/teachers/curriculuminspirations/james-tanton-biography.
ABOUT THE AUTHOR: Believing that mathematics really is accessible to all, James Tanton (PhD, Mathematics, Princeton 1994) is
committed to sharing the delight and beauty of the subject. In 2004 James founded the St. Mark’s Institute of Mathematics, an outreach program promoting joyful and effective mathematics education. He worked as a fulltime high-school teacher at St. Mark’s School in Southborough, MA (2004-2012), and he conducted, and continues to conduct, mathematics courses and workshops for mathematics teachers across the nation and overseas. James is the author of Solve This: Math Activities for Students and Clubs (MAA, 2001), The Encyclopedia of Mathematics (Facts on File, 2005), Mathematics Galore! (MAA, 2012), Geometry: An Interactive Journey to Mastery (The Great Courses, 2014), Without Words: Volumes 1 and 2
(Tarquin 2015), Trigonometry: A Clever Study Guide (MAA, 2015), and twelve self-published texts. He is the 2005 recipient of the Beckenbach Book Prize, the 2006 recipient of the Kidder Faculty Prize at St. Mark’s School, and a 2010 recipient of a Raytheon Math Hero Award for excellence in school teaching and currently serves as the Mathematician-at-Large for the Mathematical Association of America. James is the author of Edfinity’s Zen Master’s Series For Middle School Students - a unique collection of digital titles for the modern educator and student.
Edfinity, a division of Looking Glass Ventures, is an educational technology company headquartered in Silicon Valley that offers transformative educational technology solutions and digital content to educators and students worldwide. Edfinity works with the world’s premier academic associations, research organizations, and educational institutions to provide equitable access to exceptional educational content. Palo Alto | Boston http://edfinity.com Edfinity is a registered trademark of Looking Glass Ventures, LLC. All other trademarks are the property of their respective owners. Copyright 2016 Looking Glass Ventures, LLC. All rights reserved 1/16. ISBN: 978-1-944931-02-5
