CHAPTER 9: DISCRETE MATH

Transcription

1 CHAPTER 9: DISCRETE MATH (Chapter 9: Discrete Math) 9.01 Calculus tends to deal more with continuous mathematics than discrete mathematics. What is the difference? Analogies may help the most. Discrete is digital; continuous is analog. Discrete is a dripping faucet; continuous is running water. Discrete math tends to deal with things that you can list, even if the list is infinitely long. Math 45 is the Discrete Math course at Mesa. SECTION 9.1: SEQUENCES AND SERIES, and SECTION 9.6: COUNTING PRINCIPLES PART A: SEQUENCES An infinite sequence can usually be thought of as an ordered list of real numbers. It is often denoted by either: a 1, a, a 3, or a 0, a 1, a, (this is more common in Calculus) where each of the a i s represent real numbers called terms. If we re lucky, there is a nice pattern to our sequence of numbers a pattern that we can describe using simple mathematical expressions.

2 (Chapter 9: Discrete Math) 9.0 Write the first three terms of the sequence with general n th term a n = n 3 1. Assume that we begin with n = 1. Solution n = 1: a 1 = 1 n = : a = ( ) 3 1 = 0 ( ) 3 1 = 7 n = 3: a 3 = ( 3) 3 1 = 6 The first three terms are: 0, 7, and 6. We may graph terms as the points ( n, a n ), just as we used to graph points ( x, f ( x) ). Technical Note: Observe that the graph of the sequence is a discretized version of the graph of f x ( ) = x 3 1, where the domain is the set of positive integers. In this sense, sequences may be thought of as functions.

3 (Chapter 9: Discrete Math) 9.03 PART B: SIGN ALTERNATORS Alternating sequences have terms that alternate (i.e., consistently switch) between positive and negative terms. They arise frequently in Calculus. Write the first four terms of the sequence where a n = ( 1) n n. Assume that we begin with n = 1. Solution n = 1: a 1 = 1 n = : a = 1 n = 3: a 3 = 1 ( ) 1 1 = 1 ( ) = ( ) 3 3 = 3 n = 4: a 4 = ( 1) 4 4 = 4 Recall that the parity of an integer is either even or odd. The parity of n here determines the sign of the term. How can we obtain an alternating sequence like the one above, but starting with a positive term? Let s adjust our sign alternator. Try: a n = ( 1) n+1 n, or a n = ( 1) n 1 n In either case, we obtain: 1,, 3, 4, PART C: EVEN VS. ODD n yields even integers, where n Z. n 1 and n + 1 yield odd integers, where n Z. These are often used to describe sequences in Calculus.

4 (Chapter 9: Discrete Math) 9.04 PART D: FACTORIALS Do you see the n! button on your calculator? This is n factorial (not n excited). It is often used to describe sequences. We define: 0!= 1 If n is a positive integer ( n Z ) +, s n! = the product of the integers from 1 to n 0!= 1 1!= 1!= 3!= 3 4!= 4 If n is large: n!= 1 Equivalently: n! = n ( )( 1) = ( )( ) ( 1) = 6 ( )( 3) ( ) ( 1) = 4 ( )( ) ( 3) ( n) ( )( n 1) n There is a short cut: 4!= 4 3! ( ) ( )( 1) (Think: Countdown.) Factorial is at the same level as exponentiation in the order of operations; they both represent successive multiplications. Therefore, the factorial operator (!) binds more tightly than the multiplication operator lower in the order of operations. In this case, 4!= ( 4) ( 3! ); this may be clearer. ( ), which is In general, if n Z + : n!= n n 1 ( )!, and 0!= 1 This is called a recursive definition of the factorial function, because it describes how previous values determine later values. In this case, we show how we get from n 1 ( )! to n!.

5 (Chapter 9: Discrete Math) 9.05 Technical Note: Can we talk about 3.5!, say? The gamma function, which you may study in Math 151: Calculus II at Mesa, is a continuous version of the factorial function. PART E: FACTORIALS, COUNTING PRINCIPLES, AND SIMPLIFICATIONS (SECTION 9.6) How many ways are there to order a standard deck of 5 cards? Solution In general: There are 5 possibilities for the first card. Once the first card is set, there are 51 possibilities for the second card. Once the first two cards are set, there are 50 possibilities for the third card. And so on Once the first 51 cards are set, there is only 1 possibility left for the last card. By the Fundamental Counting Principle described on p.663 of Larson, we multiply together these numbers of possibilities. The number of ways is: ( 5) ( 51) ( 50) 1 That s an 8 followed by 67 digits literally astronomical! ( ), better known as 5!, which is about n! grows in value very fast as n grows. Many calculators can t even handle 100!. The number of ways to order n distinct items is: n! The different possible orders are called permutations.

6 (An elaboration on 9 on p.668 in Larson.) (Chapter 9: Discrete Math) 9.06 How many possible hands of five cards can be drawn from a deck of 5 cards? In other words, how many ways are there to choose five cards from a deck of 5? (The order among the five chosen cards does not matter.) Solution 5 This number is denoted by 5 C 5, or 5, called 5 choose 5. This is the number of combinations of 5 (distinct) items taken 5 at a time. It turns out that 5 5 = 5! 5!47!. Why? There are 5! ways to order the complete deck of 5 cards. Imagine a divider separating the first five cards from the last (other) 47 cards. Let s call the first five cards the winners and the last 47 cards the losers. We do not care about the order among the five winners, so we divide by 5!. (The technical reason for this is somewhat subtle and may not be immediately clear.) Likewise, we do not care about the order among the 47 losers (though poker players may beg to differ), so we divide by 47!, also. How can we simplify and compute without a problem? 5! 5!47! so that a calculator can handle it Good calculators should be able to handle this, but it is a good skill to know how to cancel factors when there are factorials involved. The idea here is to unravel one or more of the factorials so that we can cancel factorials. 5! ! = 5!47! 5! 47! =,598,960 By the way, there is about a chance of getting at least a pair.

7 (Chapter 9: Discrete Math) 9.07 Incidentally, we were guaranteed to get an integer because of the nature of the problem. This may not have been obvious from the form of the expression. In general: The number of ways to choose r distinct items from a set of n distinct items (if the order among the chosen items is irrelevant) is given by: C = n n r r = n! r! ( n r)! Think: The number of ways to choose r winners and, therefore, n r losers. n These r numbers are called binomial coefficients for reasons we shall see in Section 9.5. What about n P r? See if you can find the n C r button on your calculator. You may also see this button on your calculator. It is the number of partial permutations of 5 (distinct) items taken 5 at a time. Let s say we actually care about the order of the five chosen cards. The number of such partial permutations is given by: 5! ! = 47! 47! 1 = 311,875,00 The idea is that we take the number of total permutations of the 5 cards (namely, 5!), and we divide by 47!, because we do not care about the order among the losers. This time, we do not divide by 5!, because we do care here about the order among the winners.

8 PART F: RECURSIVELY DEFINED SEQUENCES Recall the recursive definition of factorials: n! = n ( n 1)! n Z + We should add that 0!= 1, because we need to know where to start. (Chapter 9: Discrete Math) 9.08 ( ) Likewise, we can define sequences by describing how terms are built upon previous terms. Perhaps the most famous recursively defined sequence is the Fibonacci sequence (named after Leonardo of Pisa, or Fibonacci, a famous Italian mathematician of the Middle Ages who has been credited with introducing Arabic numerals to the West), which is described on p.616 in Larson. There are surprising applications of this sequence in nature: sunflower petals, pine cone and seashell patterns, etc. You are virtually assured of dealing with this sequence in a Discrete Math class. Finding the general n th term expression is a bit of a challenge! It turns out to be: a n = 1 n n 5. Yes, we get integers out of this thing! Technical details and other properties are available at the URL: Consider the sequence recursively defined by: a 1 = 10 a k +1 = a k k Z + ( ) Find the first three terms of this sequence, beginning with n = 1. Solution We essentially double one term in the sequence to obtain the next term. a 1 = 10 a = a 1 = 10 a 3 = a = 0 ( ) = 0 ( ) = 40

9 (Chapter 9: Discrete Math) 9.09 PART G: SUMMATION (SIGMA) NOTATION Σ is the Greek letter known as uppercase sigma. (Lowercase sigma, σ, is famous for denoting standard deviation in statistics.) Σ is a summation operator; we use it to efficiently indicate the addition of terms. ( ) Evaluate the sum 5i 1. 3 i=0 a i Terminology and Notation Solution Here, i serves as the index of summation. It does not matter if we use j, k, etc., as long as there are no conflicts with other notations, and we use these notations consistently. We can think of i as a dummy variable. It s like your new co-worker if we replaced one dummy with another, nothing really changes. In Calculus, dummy variables are used in definite integration. Here, 0 is the lower limit (of summation), and 3 is the upper limit. We need to successively replace i with the integers from 0 through 3 and then add the results. 3 a i = a 0 + a 1 + a + a 3 i=0 ( ) 1 = a 0 = = 66 ( ) 1 a 1 ( ) 1 a ( ) 1 a 3

10 (Chapter 9: Discrete Math) 9.10 PART H: SERIES A series arises when we take the sum of terms of a sequence. The following is an example of a finite series: n k =1 a k = a 1 + a + a a n The following is an example of an infinite series: k =1 a k = a 1 + a + a 3 + In fact, the finite series above represents the n th partial sum of the infinite series above; it represents the sum of the first n terms.

11 (Chapter 9: Discrete Math) 9.11 SECTION 9.: ARITHMETIC SEQUENCES and PARTIAL SUMS PART A: WHAT IS AN ARITHMETIC SEQUENCE? The following appears to be an example of an arithmetic (stress on the me ) sequence: a 1 = a = 5 a 3 = 8 a 4 = 11 We begin with. After that, we successively add 3 to obtain the other terms of the sequence. An arithmetic sequence is determined by: Its initial term Here, it is a 1, although, in other examples, it could be a 0 or something else. Here, a 1 =. Its common difference This is denoted by d. It is the number that is always added to a previous term to obtain the following term. Here, d = 3. Observe that: d = a a 1 = a 3 a = = a k +1 a k ( k Z ) + = The following information completely determines our sequence: The sequence is arithmetic. (Initial term) a 1 = (Common difference) d = 3

12 (Chapter 9: Discrete Math) 9.1 In general, a recursive definition for an arithmetic sequence that begins with a 1 may be given by: a 1 given a k +1 = a k + d ( k 1; "k is an integer" is implied) The arithmetic sequence 5, 0, 15, 10, can be described by: a 1 = 5 d = 5

13 (Chapter 9: Discrete Math) 9.13 PART B : FORMULA FOR THE GENERAL n th TERM OF AN ARITHMETIC SEQUENCE Let s begin with a 1 and keep adding d until we obtain an expression for a n, where n Z +. a 1 = a 1 a = a 1 + d a 3 = a 1 + d a 4 = a 1 + 3d a n = a 1 + ( n 1)d The general n th term of an arithmetic sequence with initial term a 1 and common difference d is given by: a n = a 1 + ( n 1)d Think: We take n 1 steps of size d to get from a 1 to a n. Note: Observe that the expression for a n is linear in n. This reflects the fact that arithmetic sequences often arise from linear models. Find the 100 th term of the arithmetic sequence:, 5, 8, 11, (Assume that is the first term. ) Solution a n = a 1 + ( n 1)d a 100 = = + 99 = 99 ( )( 3) ( )( 3)

14 PART C : FORMULA FOR THE n th PARTIAL SUM OF AN ARITHMETIC SEQUENCE (Chapter 9: Discrete Math) 9.14 The n th partial sum of an arithmetic sequence with initial term a 1 and common difference d is given by: S n = n a + a 1 n Think: The (cumulative) sum of the first n terms of an arithmetic sequence is given by the number of terms involved times the average of the first and last terms. Find the 100 th partial sum of the arithmetic sequence:, 5, 8, 11, Solution We found in the previous that: a 100 = 99 S n = n a + a 1 n + 99 S 100 = ( 100) 301 = ( 100) = 15,050 i.e., = 15,050 This is much easier than doing things brute force on your calculator! Read the Historical Note on p.68 in Larson for the story of how Gauss quickly 100 computed the sum of the first 100 positive integers, k = Use our formula to confirm his result. Gauss s trick is actually used in the proof of our formula; see p.694 in Larson. We will touch on a related question in Section 9.4. k =1

15 (Chapter 9: Discrete Math) 9.15 SECTION 9.3: GEOMETRIC SEQUENCES, PARTIAL SUMS, and SERIES PART A: WHAT IS A GEOMETRIC SEQUENCE? The following appears to be an example of a geometric sequence: a 1 = a = 6 a 3 = 18 a 4 = 54 We begin with. After that, we successively multiply by 3 to obtain the other terms of the sequence. Recall that, for an arithmetic sequence, we successively add. A geometric sequence is determined by: Its initial term Here, it is a 1, although, in other examples, it could be a 0 or something else. Here, a 1 =. Its common ratio This is denoted by r. It is the number that we always multiply the previous term by to obtain the following term. Here, r = 3. Observe that: r = a = a 3 = = a k +1 ( k Z ) + = a 1 a a k The following information completely determines our sequence: The sequence is geometric. (Initial term) a 1 = (Common ratio) r = 3

16 (Chapter 9: Discrete Math) 9.16 In general, a recursive definition for a geometric sequence that begins with a 1 may be given by: a 1 given a k +1 = a k r ( k 1; "k is an integer" is implied) We assume a 1 0 and r 0. The geometric sequence, 6, 18, 54, can be described by: a 1 = r = 3

17 (Chapter 9: Discrete Math) 9.17 PART B : FORMULA FOR THE GENERAL n th TERM OF A GEOMETRIC SEQUENCE Let s begin with a 1 and keep multiplying by r until we obtain an expression for a n, where n Z +. a 1 = a 1 a = a 1 r a 3 = a 1 r a 4 = a 1 r 3 a n = a 1 r n1 The general n th term of a geometric sequence with initial term a 1 and common ratio r is given by: a n = a 1 r n1 Think: As with arithmetic sequences, we take n 1 steps to get from a 1 to a n. Note: Observe that the expression for a n is exponential in n. This reflects the fact that geometric sequences often arise from exponential models, for example those involving compound interest or population growth.

18 (Chapter 9: Discrete Math) 9.18 Find the 6 th term of the geometric sequence:, 1, 1, (Assume that is the first term. ) Solution Here, a 1 = and r = 1. a n = a 1 r n1 a 6 = ( ) 1 = = ( ) 1 ( ) 1 = Observe that, as n, the terms of this sequence approach 0. ( ) 1 < r < 1 Assume a 1 0. Then, a 1 r n1 0 as n ( ) i.e., r <1

19 (Chapter 9: Discrete Math) 9.19 PART C : FORMULA FOR THE n th PARTIAL SUM OF A GEOMETRIC SEQUENCE The n th partial sum of a geometric sequence with initial term a 1 and common ratio r (where r 1) is given by: S n = a a r n r n or a 1 r 1 1 r You should get used to summation notation: Remember that S n for a sequence starting with a 1 is given by: n S n = a k k =1 = a 1 + a + + a n Because a k = a 1 r k 1 for our geometric series: n S n = a 1 r k 1 = a 1 + a 1 r + a r a r n1 1 1 k =1 a a 3 a n = a 1 a 1 r n 1 r ( according to our theorem in the box above) Note: The book Concrete Mathematics by Graham, Knuth, and Patashnik suggests a way to remember the numerator: first in first out. This is because a 1 is the first term included in the sum, while a 1 r n is the first term in the corresponding infinite geometric series that is excluded from the sum. Technical Note: The key is that 1+ r + r + + r n1 = 1 r n. You can see that this 1 r is true by multiplying both sides by ( 1 r). Also see the proof on p.694 of Larson. Technical Note: If r = 1, then we are dealing with a constant sequence and essentially a multiplication problem. For example, the 4 th partial sum of the series is = ( 4) ( 7)= 8. In general, the n th partial sum of the series a 1 + a 1 + a is given by na 1.

20 (Chapter 9: Discrete Math) 9.0 Find the 6 th partial sum of the geometric sequence, 1, 1, Solution Recall that a 1 = and r = 1 for this sequence. We found in the previous that: a 6 = 1 16 We will use our formula to evaluate: S 6 = Using our formula directly: S n = a 1 a 1 r n 1 r 1 r n or a 1 1 r If we use the second version on the right 1 r n S n = a 1 1 r S 6 = =

21 (Chapter 9: Discrete Math) 9.1 = = = = We can also use the first version and the first in first out idea: S 6 = First out is: a 7 = 1 3 S n = a a r n r 1 S 6 = = = =

22 (Chapter 9: Discrete Math) 9. PART D: INFINITE GEOMETRIC SERIES An infinite series converges (i.e., has a sum) The S n partial sums approach a real number ( as n ), which is then called the sum of the series. In other words, if lim n S n = S, where S is a real number, then S is the sum of the series. Consider the geometric series: Let s take a look at the partial (cumulative) sums: S 1 = 1 S = 3 4 S 3 = 7 8 S 4 = It appears that the partial sums are approaching 1. In fact, they are; we will have a formula for this. This series has a sum, and it is 1. The figure below may make you a believer:

23 (Chapter 9: Discrete Math) 9.3 The geometric series has no sum, because: lim n S n = The geometric series has no sum, because the partial sums do not approach a single real number. Observe: S 1 =1 S =0 S 3 =1 S 4 =0 ( ) An infinite geometric series converges 1 < r < 1 i.e., r <1 Take another look at the s of this Part. It is true that an infinite geometric series converges Its terms approach 0. Warning: However, this cannot be said about series in general. For example, the 1 famous harmonic series = 1+ 1 k =1 k does not converge, even though the 3 terms of the series approach 0. In order for a series to converge, it is necessary but not sufficient for the terms to approach 0. No infinite arithmetic sequence (such as ) can have a sum, unless you include as an arithmetic sequence.

24 (Chapter 9: Discrete Math) 9.4 The sum of a convergent infinite geometric series with initial term a 1 and common ratio r, where 1 < r < 1, is given by: i.e., r <1 S = a 1 1 r Technical Note: This comes from our partial sum formula S n = a 1 a 1 r n ( ) if 1 < r < 1 fact that a 1 r n 0 as n 0. i.e., r <1 1 r and the Write 0.81 as a nice (simplified) fraction of the form integer integer. Recall how the repeating bar works: 0.81 = Note: In Arithmetic, you learned how to use long division to express a nice fraction as a repeated decimal; remember that rational numbers can always be expressed as either a terminating or a nicely repeating decimal. Now, after all this time, you will learn how to do the reverse! Solution 0.81 can be written as: Observe that this is a geometric series with initial term a 1 = 0.81 and common ratio r = = 1 = 0.01; because r < 1, the series 100 converges. The sum of the series is given by: S = a 1 1 r = = = = 9 11

25 (Chapter 9: Discrete Math) 9.5 Again, you should get used to summation notation: S = k =1 a 1 r k 1 ( )( 0.01) k 1 = 0.81 k =1 = 9 11 If you make the substitution i = k 1, the summation form can be rewritten as: S = 0.81 k =1 ( )( 0.01) k 1 ( )( 0.01) i = 0.81 i=0 In Calculus, 0 is more common than 1 as a lower limit of summation.

26 (Chapter 9: Discrete Math) 9.6 SECTION 9.4: MATHEMATICAL INDUCTION What is the sum of the first n positive integers, where n Z +? In other words, what is n? According to our formula for the n th partial sum of an arithmetic sequence (see Section 9.), the answer is: S n = n 1+ n = n ( n + 1 ) Handshake Problem (Cool, but Optional) There s another way of seeing why this formula works out. Imagine n + 1 people walking into a room one-by-one. Whenever a person walks into a room, he/she must shake hands exactly once with every other person who is currently in the room, and those are the only handshakes they make. The first person who walks into the room shakes nobody s hand, the second person shakes the first person s hand, the third person shakes the first two people s hands, and so on, until the last person shakes the other n people s hands. This means that every distinct pair of people eventually shake hands exactly once. The total number of handshakes then equals the number of distinct pairs of people that can be formed from a group of n + 1 people. We see that the number of handshakes equals both n and n + 1 ( = n + 1)! (! (( n + 1) )! = n + 1)! ( n + 1)n ( n 1)!! ( n 1)! = = n ( n + 1 ). ( n 1)! ( ) Therefore: n = n n + 1 In this section, we will use mathematical induction to prove that: n = n ( n + 1 ) Mathematical induction is used to prove conjectures, statements that we believe to be true but are as yet unproven. Unfortunately, the method of mathematical induction cannot give us a formula such as n = n ( n + 1 ) ; we must first guess at such a formula by perhaps working out a few cases and using trial-and-error. (For example, try out Problem #90 in Section 9. on p.633 in Larson.) Induction is then used to verify our guess.

27 (Chapter 9: Discrete Math) 9.7 Induction is commonly used in Discrete Math and in Linear Algebra. It is even used in continuous mathematics; in Calculus, for example, induction is used to prove integration formulas involving powers of (or products of powers of) trig functions. The Domino Image Visualize a half-line (or a ray ) of infinitely many dominoes. When are we guaranteed that all the dominoes will eventually fall? If we are guaranteed the following: The first domino falls. (This corresponds to the Basis Step.) The fall of one domino guarantees the fall of the next domino. (This corresponds to the Inductive Step.) The Proof of Our Formula Conjecture: Basis Step: The following is true for all positive integers n ( i.e., n Z + ): ( ) P n, which states that: n = n n + 1 Verify that P 1 is true. (i.e., Verify that P n is true for n = 1.) 1 = 1 ( 1+ 1 ) ( ) 1 = 1 1 = 1 This demonstrates that P 1 is true, because we have shown that P 1 is equivalent to the true statement 1 = 1.

28 (Chapter 9: Discrete Math) 9.8 Note: If we only had to prove P n for all integers n such that n 7, say, then we would verify P 7 as our Basis Step. Inductive Step: Let k be any fixed positive integer. (Some books use n; whatever you do, be consistent throughout the problem!) Assume that P k is true. (This assumption is called the Inductive Hypothesis, which we will abbreviate as I.H.) Using the Domino Image: Assume that the k th domino has fallen. Here, we assume: k = k ( k + 1 ) Show that P k +1 is then true. Using the Domino Image: We must then show that the ( k + 1) st domino must then fall as a result. Here, we must show that, as a result: ( ) ( ( k + 1) = k + 1 ) ( k + 1) + 1, or ( k + 1) ( k + ) This is somewhat similar to a verification problem in trig in that the righthand side is something of a TARGET.

29 k + 1 ( ) = k + k + 1 Apply the I.H. to this. ( ) = k k + 1 = k k + 1 = k k + 1 ( ) ( ) ( ) + k k + 1 ( ) + ( k + 1) ( )( k + 1) = k + = k + 1 ( )( k + ) (Chapter 9: Discrete Math) 9.9 ( ) ( from Factoring) ( TARGET) Note: The argument above works even for k = 1, k =, and k = 3, even though the first line may seem incompatible with those cases. Conclusion We may now write: or P n is true for all positive integers n. P n is true for all positive integers n. QED. Note: QED stands for the Latin phrase Quod Erat Demonstrandum. It effectively means that which was to have been proven. Read p.648 in Larson for formulas for sums of powers of the first n positive integers. Challenge Problem Use induction to prove that every amount of postage greater than or equal to 1 cents can be formed by using just 4- and 5-cent stamps.

30 SECTION 9.5: THE BINOMIAL THEOREM (Chapter 9: Discrete Math) 9.30 PART A: INTRO How do we expand ( a + b) n, where n is a nonnegative integer? ( a + b) 0 =1 ( a + b) 1 = a + b ( a + b) = a + ab + b ( a + b) 3 = ( a + b) ( a + b) ( a + b) 10 = YUK! Do first = a 3 + 3a b + 3ab + b 3 Look for patterns Take ( a + b) 3. Let s look at the variable parts of the terms. ( a + b) 3 = a 3 + 3a b + 3ab + b 3 Starts with a 3 Ends with b 3 At each step, the exponent on a by 1 the exponent on b by 1 a 0 = 1, b 0 = 1, so they are invisible. Each term has degree 3.

31 (Chapter 9: Discrete Math) 9.31 In general, ( a + b) n = a n + + b n (n is a whole number) ( n+1) terms (# terms = power + 1) What about the coefficients? They are given by PART B: PASCAL S TRIANGLE The ingredients: Tent of 1 s Any other entry = sum of the two entries immediately above Row n gives the coefficients for ( a + b) n. ( a + b) 0 = 1 ( a + b) 1 = a + b ( a + b) = a + ab + b ( a + b) 3 = a 3 + 3a b + 3ab + b 3 ( a + b) 4 = a 4 + 4a 3 b + 6a b + 4ab 3 + b 4 (See my website for some magical properties of Pascal s triangle!)

32 PART C: EXPANDING POWERS OF GENERAL BINOMIALS Expand and simplify ( 3x y) 3 using the Binomial Theorem. Solution 3 We will use the template for a + b. ( 3x y) 3 = 3x + ( y) Sub a= 3 x Sub b = y [ ] 3 = a + b 3 = a 3 + 3a b + 3ab + b 3 Sub back: a ( 3x), b y ( ) ( ) 3 + 3( 3x ) ( y) + 3( 3x) ( y) + ( y) 3 = 3x Simplify. First, do powers. = 7x x ( ) y ( ) + 3( 3x) y = 7x 3 7x y + 9xy y 3 ( ) y 3 (Chapter 9: Discrete Math) 9.3

33 PASCAL'S TRIANGLE THE MAGIC OF PASCAL'S TRIANGLE Ên This represents a way to write down the "early" binomial coefficients Á ˆ easily. Ër Each row begins and ends with "1". (We have a "tent" of "1"s.) Every other entry equals the sum of the two entries immediately above it. Here it is: 1 Row 0: Contains 1 1 Row 1: Contains 1 1 Row : Contains Row 3: Contains Row 4: Contains Ê0 Á ˆ Ë0 Ê1 1 Á ˆ, Ê Á ˆ Ë0 Ë1 Ê Á ˆ, Ê Á ˆ, Ê Á ˆ Ë0 Ë1 Ë Ê Á ˆ, Ê Á ˆ, Ê Á ˆ, Ê Á ˆ Ë0 Ë1 Ë Ë3 Ê Á ˆ, Ê Á ˆ, Ê Á ˆ, Ê Á ˆ, Ê Á ˆ Ë0 Ë 1 Ë Ë3 Ë4 etc. - The "histograms" of the rows approach a bell-shaped "normal" distribution! We can observe some basic properties of binomial coefficients: Symmetry about the center: Ên n Á ˆ Ê ˆ = Á Ër Ën- r (The process of choosing r winners is equivalent to the process of choosing n-r losers.) The "tent" of "1"s: Ên n Á ˆ = Ê 1 Ë0 Ë Á ˆ = n The "inner tent" of natural numbers: Ên n Á ˆ Ê ˆ = Á = n Ë1 Ën - 1 (There are n ways to get one winner and n-1 losers from a group of n people.)

34 THE PLINKO / PACHINKO APPROACH TO PASCAL'S TRIANGLE In the game of Plinko on the CBS game show "The Price is Right", contestants won money by standing over a board full of pins and dropping chips. The chips then fell into bins at the bottom of the board; each bin was labeled with a dollar amount that was added to the contestant's winnings. Now imagine a pinboard shaped like Pascal's triangle: Player 1 After pin is hit, the chip can move left or right. 1 1 After a pin is hit, the chip can move left or right. 1 1 After a pin is hit, the chip can move left or right After a pin is hit, the chip can move left or right After a pin is hit, the chip can move left or right. It turns out that the number of ways to hit a pin is equal to the number on the pin! There are 3 ways to hit the boldfaced pin labeled "3" above. One way: LLR Left (L) - first "drop" Left (L) - second "drop" Right (R) - third "drop" A second way: LRL Left (L) Right (R) Left (L)

35 A third way: RLL Right (R) Left (L) Left (L) The only way to reach the "3" is if there are exactly one "R" and two "L"s in any order. How many ways are there to choose exactly one of the three Ê3 drops to be an "R"? Á ˆ, which the 3 represents! Ë1 Likewise, there are 6 ways to hit the "6" pin, because out of four drops, we Ê4 need exactly two of them to be "R"s. There are Á ˆ = 6ways to do this. Ë Remember the construction of Pascal's triangle: Each row begins and ends with "1". (We have a "tent" of "1"s.) Every other entry equals the sum of the two entries immediately above it. How does the Plinko model exhibit this? There is 1 way to hit the "1" (all "L"s). Then, you can move right to hit the "4". There are 3 ways to hit the "3". Then, you can move left to hit the "4". So, there are 4 total ways to hit the "4". Note: If you go to a science museum that has a "Plinko-type" board with numerous balls being dropped from the top-center, the bottom of the board should look like a bell curve!

36 EXPANDING POWERS OF BINOMIALS: THE BINOMIAL THEOREM (MATH 96) Remember that Row n of Pascal's Triangle provides the coefficients for the expansion of ( a+ b) n : n n n n n a b a n a n b a n b n b n ( + ) = Ê Ë Á ˆ + Ê Ë Á ˆ -1 + Ê Ë Á ˆ Ê Ë Á ˆ 0 1 (a+b) 3 = 1a + 3a b+ 3ab + 1b 3 3 Here's why this expansion of a+ b ( ) 3 is correct: You can start by writing down 8 terms, each corresponding to a sequence of choices. 3 ( a+ b) = ( a+ b) ( a+ b) ( a+ b) Choose " a" or " b" " Choose a" or " b" " Choose a" or " b" ÏÊ3 Á ˆ 3 Ô =1 way to never ( a+ b) = aaa { ÌË0 3 a Ô Óchoose " b" ÏÊ3 ÔÁ ˆ =3 ways to choose " b" + aab + aba + baa ÌË a b Ô Óexactly once among the 3 factors ÏÊ3 ÔÁ ˆ =3 ways to choose " b" + abb + bab + bba ÌË ab Ô Óexactly times among the 3 factors + bbb 13 3 b = a + 3a b+ 3ab + b 3 3 ÏÊ3 ÔÁ ˆ =1 way to always ÌË3 Ô Óchoose " b" Notice that choosing the "a" or the "b" from each factor is analogous to dropping "left" or "right" at each step in our Plinko model!