### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Study Guides for Discrete Math 2405 2405

UHD

### View Full Document

## About this Document

## 60

## 0

## Popular in Discrete Mathematics

## Popular in Mathematics (M)

This 38 page Study Guide was uploaded by BigBoss on Thursday December 31, 2015. The Study Guide belongs to 2405 at University of Houston Downtown taught by Dr. Linda Becerra in Summer 2015. Since its upload, it has received 60 views. For similar materials see Discrete Mathematics in Mathematics (M) at University of Houston Downtown.

## Reviews for Study Guides for Discrete Math 2405

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 12/31/15

Name__________________________________ Test 1study guide Chapter 1.1, 1.3 Tables that you might need to study Make a truth table for the following propositions a. ¬ (¬ p v (p v q)) q (Is this a tautology?) p q ¬p (p v q) ¬ (¬ p v (p v q)) ¬ (¬ p v (p v q)) q b. (¬ p ^ (p q)) ¬q (Is this a tautology?) p q (p q) ¬q (¬ p ^ (p q)) (¬ p ^ (p q)) ¬q 1 c. ¬ p (q r) ≡ (p v r) (Are they equal?) p q r (q r) (p v r ) ¬p (q r) q (p v r) Implications ( ) p q q p Converse ¬p ¬q inverse ¬q ¬p contrapositive Write the converse, inverse and contrapositive of the given implication: If cows eat grass, then 2+3 = 4 Proposition True or False? Converse Inverse Contrapositive If horses fly, then sheep roam Proposition True or False? Converse Inverse Contrapositive If hats are worn on heads, then two Proposition True or False? sandwiches can form scrap metal Converse Inverse Contrapositive 2 State the law used in the identity i. ¬ ( p ^ q ) ν T ≡ T _________________________________ ii. T ν ¬ (¬ p ^ q ) ≡ ¬ (¬ p ^ q ) _________________________________ iii. ¬ (¬ p ^ q ) v F ≡ ¬( ¬ p ^ q ) _________________________________ iv. ¬ ( p ^ q ) ^ T ≡ ¬ ¬ p v ¬q _________________________________ v. p ^ (r v q ) ≡ ( r v q ) ^ p _________________________________ vi. (p ^ r ) v (p ^ q ) ≡ p ^ (r v q ) _________________________________ vii. ¬ p ^ ¬ (r ^ q ) ≡ ¬ (p v ( r v q )_________________________________ viii. (p v r ) q ≡ ¬ (p v r ) v q _________________________________ ix. (r v q ) v ¬ (r v q ) ≡ T _________________________________ x. [ ¬ ( ¬ r v ¬q ) (r v q )] v T ≡ _________________________________ Show that ¬ (¬ p v (p v q)) q is a tautology Proof: ≡ ____________________________________ by _________________________ ≡ ____________________________________ by _________________________ ≡ ____________________________________ by _________________________ ≡ ____________________________________ by _________________________ ≡ ____________________________________ by __________________________ ≡ ____________________________________ by __________________________ ≡ ____________________________________ by __________________________ ≡ ____________________________________ by __________________________ □ 3 Show that ¬ (p ^ q) v (p ^ q) ≡ T Proof: ≡ ____________________________________ by _________________________ ≡ ____________________________________ by _________________________ ≡ ____________________________________ by _________________________ ≡ ____________________________________ by _________________________ ≡ ____________________________________ by __________________________ □ Show that ¬ p (q r) ≡ q (p v r ) ≡ ____________________________________ by _________________________ ≡ ____________________________________ by _________________________ ≡ ____________________________________ by _________________________ ≡ ____________________________________ by _________________________ ≡ ____________________________________ by __________________________ ≡ ____________________________________ by __________________________ □ 4 Show that p (q r) ≡ q (p v r) ≡ ____________________________________ by __________________________ ≡ ____________________________________ by __________________________ ≡ ____________________________________ by __________________________ ≡ ____________________________________ by __________________________ □ Show that(¬ p ^ (p q)) ¬ q Proof: ≡ ____________________________________ by _________________________ ≡ ____________________________________ by _________________________ ≡ ____________________________________ by _________________________ ≡ ____________________________________ by _________________________ ≡ ____________________________________ by __________________________ ≡ ____________________________________ by __________________________ ≡ ____________________________________ by __________________________ ≡ ____________________________________ by __________________________ ≡ ____________________________________ by __________________________ □ 5 Warning: only do this if you’re hardcore (14 steps) *It’s dangerous to go alone! Take this * Show that ¬ (p ↔ q) ≡ p ↔ ¬q Proof: 6 1.4 predicates and quantifiers ∀x P(x) A ↷ ∀ Flip upside-down ↑ “All”, “for every”, “each” universal quantifier ∃x P(x) E ↔ ∃ Reflect horizontally ↑ “There exists”, “there is”, “at least one” existential quantifier P(x) {Statement} Universal Truth value Truth value disclosure of P(x)∃x P(x) ∀x P(x) x = x + 3 All real numbers X = x + 0 All Real Numbers 0/ x = 0 Integers 0 – x = x Positive integers 0 – x = x Natural numbers What is the truth value of [∀x P(x) ∃_______________________________________________ What is the truth value of [∃x P(x) ∀________________________________________________ 1.5 Nested quantifiers Example 1: Assume that the universe of discloser for variables x and y consists on all real numbers i. Translate ∃x ∀y (x + y = 0 ) into an English statement ii. What is the truth value of ∃x ∀y (x + y = 0, why? 7 2.1 sets All sets are used in CAPITAL letters A , B , C X … Two sets are equal ↔ if the elements are the same, order does not matter R = {x | x is a real number} “|” means “such that” Universal set is normally denoted as “U” which is everything not in the set Subsets Sets A is saidto be a subset of set B if and only if every element in A is also an element of B. A ⊆ B ↔ ∀ ̓ x, x € A x € B ↑ A is a subset of B If and only if all elements x that are in set A is .lso in set B Picture of ⊆ B Example: Let A = {a, b, c}, B = {a, b, c, d} C = {b, c, d} A B ⊂ Set of or = C B ⊆ Subset of A C D B D A Cardinality The amount of elements that are in a set 8 Find the cardinality of the following sets i. |{1000}| = __________________________________ ii. |{∞}| = __________________________________ iii. |{1, 2, 3, …. ,1000}| = __________________________________ iv. |{1, 2, 3, 1000}| = ___________________________________ Power set The set of |P (A)| contains 2 , where n is the amount of elements in the set Find the power set: a. P(ø) = b. P(z, x ,y ) = c. P(P(h, a, x) = d. P (P(P( ß))) = Let A be the universal set Proposition True/False Ø ⊆ A { Ø} ⊆ A Ø ⊆ P(A) {Ø} ⊆ P(A) Ø ∈ A Let A = {a, b, c} Proposition True/False {a} ∈ A {a} ⊆ A {a} ∈ P(A) {a} ⊆ P(A) {{a}} ⊆ P(A) 9 Cartesian product The set of all ordered pairs (a,a∈)A and b∈ B, denoted as (A X B) Find the Cartesian product of each set a. A = { a, b } , B = { 1, 2, 3} A X B = B X A = b. C = {P, C} , D = { M, A , C} C X D = D X C = c. Z = {♠,♣, ♠, ♦} , Y = { β, λ, μ} , W= { Ω, ψ } W x Z x Y = Y x Z x W = Z x Y x W = Y x W x Z = W x Y x Z = 10 2.2 Set operations Union of sets A ∪ B A ∪ B = { x |( x ∈ A) v ( x ∈ B) } ↑ All elements are in both A or B Draw the Venn diagram: Intersection of setsA ∩ B A ∩ B = { x | ( x ∈ A ) ^ ( x ∈ B)} ↑ All elements that are both in A and B are the same Draw the Venn diagram: Disjoint setsA ∩ B = ø The sets are disjoint only if A and B have nothing in common Draw the Venn diagram: 11 Difference of sets A– B A – B ={x | (x ∈ A) ^(x ∉ B) } ↑ Some elements are in A but some elements are not in B Draw the Venn diagram: Complement of set A A = {x | x ∉A} ↑ All elements that are not in A, but that are in the universal set U Draw the Venn diagram: Example: Now draw a Venn diagram (A – B) – C A – (B – C) 12 Name____________________________________________________ Test2studyguide Chapter 1.7 Proofs: Proofs are normally follow pq Direct proof Indirect proof Proof by contradiction Begins by assuming p is true Begins by assuming the negation of Begins by assuming p^ ¬q is true q is a true statement Proven by using: Proven by using: Proven by using: Logic, definitions, axioms, Logic, definitions, axioms, Logic, definitions, axioms, theorems, theorems, and simple algebra. theorems, and simple algebra. and simple algebra. Until q is concluded Until the negation of p is concluded Until a contradiction is concluded (pq) (¬q ¬p) (TF ≡ F ) Proofs that are interesting but you may not need Vacuous proof: an implication when the hypothesis of the implication is always false Example: x = -x Trivial proofs: whenever the conclusion of an implication is always true Conjecture: a mathematical proposition whose truth value is unknown Axioms – a statement that is established as a true statement The following axioms are vital for building proofs: Closure of addition: If a and b are integers, then a +b is an integer (used whenever you add in the algebra step) Closure of multiplication: If a and b are integers, then a * b is an integer (used whenever you multiply in the algebra step) Vital definitions: (These proofs are the most common in direct proofs) Definition of even: An integer n is even if there exists an integer k such that n = 2k Definition of odd: An integer n is odd if there exists an integer k such that n = 2k +1 Lemma: A theorem that is used within a proof to assist in solving the proof How to build a proof: Follow these steps and you will be a pro! For direct Proofs: Step 1: Identify the hypothesis “p” and the conclusion “q”, then find how pq -Implication is commonly denoted as “if, then” Step 2: Once p and q is identified, then use definitions to establish the variables that are needed Example: “By the definition of even an integer n is even if there exists an integer k such that n=2k” Step 3: The “meat” of the problem, use algebra, or any other means to prove that p q 2 Example: The algebraic step for If n is even, then n is even (using substitution from the definition of even) 2 n (2k) n statement to start off with = (2k) substitution of n for the definition of even = 2(2k ) proving that n is even Step 4: Closure of integers (axioms) - state the arithmetic operation(s) used to solve the “meat” of the problem Example: (following up rom step 3) “Since 2 and k are integers and integers have closure with respect to MULTIPLICATION, then 2k 2 is an integer…” Step 5: Conclusion of the proof, end the proof by re-stating pq. Example: (following up from step 4) 2 “Thus by definition of even n is even when n is assumed Q.E.D. “ Now with the same steps, prove the following lemma, 2 Lemma 1: If n is even, then n is even. Proof: For contrapositive proofs: Identical to solving direct proofs, but invert the hypothesis and conclusion Step 1: Identify the hypothesis “p” and the conclusion “q”, then negate the hypotheses and the conclusion, then switch the order (¬q ¬p) Step 2: Once ¬p and ¬q is identified, then use definitions to establish the variables that are needed Step 3: The “meat” of the problem, use algebra, or any other means to prove that ¬q ¬p Step 4: Closure of integers (axioms) - state the arithmetic operation(s) used to solve the “meat” of the problem Step 5: Conclusion of the proof, end the proof by re-stating ¬q¬p. Now prove by contrapositive If x+ y ≥ 2, where x and y are real numbers, then x ≥ 1 or y ≥ 1. Proof: Assume _____________________ or ___________________. Then _____________ and _________ By using simple algebra we get that, Since _________, ___________ and __________ are integers and the integers have closure with respect to _________________________. Hence ____________________________ Q.E.D. For proofs that use contradiction Step 1: Identify the hypothesis “p” and the conclusion “q”, then identify a way to contradict the statement Step 2: Once p and q is identified, then use definitions to establish the variables that are needed Step 3: The “meat” of the problem, use algebra, or any other means to prove that p ^ ¬q Step 4: Closure of integers (axioms) - state the arithmetic operation(s) used to solve the “meat” of the problem Step 5: Conclusion of the proof, end the proof by stating the contradiction of the proof. Prove by way of contradiction (An example for you to look at) Prove that √ is irrational Proof: Assume by way of contradiction that 2√can be represented as a quotient of two integers p/q where q is not zero. Further, we assume that p/q is in lowest terms, assume that The integers p and q have no common factor. (1) ???? Thus, by assumption √ 2 = ???? now squaring both sides we get that ????2 2 = 2 or p = 2q 2 (2) ???? 2 This implies that p is even, and by theorem 1 , p must also be even. So p =2k for some integer k. Substitute into the second equation of (2) and by cancelation we see that 2 2 q = 2k (3) 2 This says that q is even and again by Theorem 1, q must also must also be even. From statements (2) and (3), it follows that p and q have both 2 as a common factor. (4) Statements (1) and (4) are contradictory. Thus√2 is not rational. Q.E.D By using contradiction prove that If n is an integer and 3n +2 is even, then n is even Proof: Chapter 2.3 Functions: Definition of a function: (A) and (B) are two different sets (they contain anything in the universe). A function “f” from A to B is an assignment of exactly one element of B to each element of A, since the function is from A to B then it is written as f: AB. A Function of A to B B One to one illustration Illustration of Onto • Definitions (for functions f over numbers): – f is strictly (or monotonically) increasing iff x>y f(x)>f(y) for all x,y in domain; – f is strictly (or monotonically) decreasing iff x>y f(x)<f(y) for all x,y in domain; • If f is either strictly increasing or strictly decreasing, then f is one-to-one. 3 – e.g. f(x)=x x (“floor of x”) is the largest integer x. x (“ceiling of x”) is the smallest integer x. Floor = 2 Ceiling = 3 Chapter 2.4 Sequences and summations Sequence: The order of numbers that is a asna notation Find the following sequences for the given functions a 0 a1 a 2 a4 {4 } { n/2 } {n!} Special integer sequences Arithmetic sequences: for sequences that are consecutive for the nth term the formula is a +d(n-1) Where ‘a’ is the first term, and “d” is the constant difference. Example: 1, 3, 5, 7, 9…. Find the nth term (general rule) ____________________________ Find the 100 term _____________________________________ Example: 0, 50, 100, 150, 200….. Find the nth term ________________________________________ Find the 100 term _______________________________________ n-1 Geometric sequences: terms that are consecutive rations which are constant general rule is ar Example: 3, 6, 12, 24, 48… Find the nth term ________________________________________ Find the 100 term _______________________________________ Example: 10, 100, 1000, 10000, 100000 Find the nth term ________________________________________ Find the 100 term _______________________________________ Summations: The sum of set of numbers (n = 1, 2, 3, 4…) What is the value of the following summation? 7 ∑???? 2 ????=5 4 ∑2 ???? ????=0 5 ∑2 (????−1) ????=1 Double summations: A summation within a summation What is the value of the following summations? ∑3 ∑ 2 ???????? = ????=1 ????=0 5 4 ∑∑ (???? + ???? ) = ????=2 ????=3 4 6 ???? ???? ∑∑ (6 + 4 ) = ????=2 ????=2 3 4 ∑∑ ???? − ???? = ????=1 ????=0 7 3 8???? − 4???? ∑∑ 2 = ????=4 ????=0 Chapter 4.1 Divisibility and modular arithmetic Def. If a and b are integers with a ≠ 0, then a divides b (a |b) and if there is an integer c such that b = ac. a | b = b/a When a | b, then a is a factor of b and b is a multiple of a True or False? 2 | 6 12 | 6 2 |6 = 3 7 is a multiple of 35 35 is a multiple of 7 Theorem 1: Let a, b, and c be integers such that a and b are not zero. Then If a | b and a |c, then a | (b +c) Proof: Division algorithm Let a be an integer and d a positive integer. Then there are unique integers q and r with 0 ≤ r < d, such that a =dq +r. Where d is the divisor, a is the dividend and q is the remainder Modular arithmetic Def. a mod b is shorthand for division remainder, a/b = (remainder of a and b) Find the remainders (notice the pattern) 9 mod 5 = 8 mod 5 = 7 mod 5= 6 mod 5 = 5 mod 5 = 4 mod 5 = 3 mod 5 = 2 mod 5 = 1 mod 5 = 0 mod 5 = -1 mod 5 -2 mod 5 = -3 mod 6 = -4 mod 8 = -6 mod 5 = -8 mod 2 = 14 mod 3 = -5 mod 2 = 36 mod 8 = -112 mod 10 = 987 mod 20 = 725 mod 35 = -65 mod 7 = -80 mod 4 = 90 mod 89 = Theorem 6 Let m>1 be a positive integer. The integers a and b are congruent modulo m if and only if there is an integer k such that a = b +km. Proof: Assume __________________________. By definition of congruency mod m, ___________________. Now by definition of divisibility, there is an integer k such that (_____________) = ______________ which by simple algebra yields _________________ for some integer k. Thus a (congruent) b mod m such that there is an integer k By simple algebra there is a n integer ______________________ which by definition of divisibility implies m|(a-b). By definition of _____mod m __________. Hence a (congruency) b if and only if there is an integer k such that a = b +km. Q.E.D. Theorem 7. Let m >1 be a positive integer. If a ≡ mod m and c ≡ d mod m, then a +c ≡ (b+ d)(mod m). Proof: Assume ___________________ and ____________________. By definition of __________________ mod m, m|__________ and m|____________, which by definition of divisibility yields. a - b =_____________ and c-d =___________ and for k and d integers a+ c-(b+ d)_ = (a+ c) –( b-d) = a-b + b –c = mk +md =m (k +d) Since k and d are integers and the integers have closure with respect to addition k + d is an integer so by definition of divisibility we set m|(a+ c) – (b+ d). Now by definition of congruency mod m, a + c ≡ (b +d) (mod m) Q.E.D. The Euclidean algorithm – exactly the same as the division algorithm ONLY for GCD Let “a” be an integer and d a positive integer. Then there are unique integers q and r with 0 ≤ r < d, such that a =dq +r. Where d is the divisor, a is the dividend and q is the remainder Example: Find the gcd of 414 and 662 with the division algorithm 662 = 414 (1) + 248 414 = 248 (1) + 166 248 = 166 (1) + 82 166 = 82 (2) + 2 2 is the gcd 82 = 2 (41) +0 Use the Euclidian algorithm to solve the gcd for the following values GCD (567, 90) GCD (47, 39) GCD (88, 234) GCD (124, 750) GCD (876, 453) Chapter 4.3 Primes and the GCD The fundamental theorem of arithmetic – any positive integer greater than one can be written as the product of primes, where primes are always increasing Examples: 2 2 100 = 2 * 5 1024 = 2 10 840 = _______________ 690 =_______________ 2048 =_______________ 999 = ________________ 567 =_________________ Greatest common denominator Def Let a and b be non-zero integers, the largest integer d such that d| a and d |b is called the greatest common divisor of a and b, denoted as gcd (a, b). To find the gcd, compare the exponents of the numbers and the maximum exponent of the base number will be used for the gcd Example: Find the gcd of 120 = 2 * 3 * 5 and 500 = 2 * 5 3 GCD (120,500) = 2 min (3,* 3 min (1,* 5min (1, 3) = 2 * 3 * 5 = 20 Find the GCD: GCD (830, 127) = GCD (56, 23) = GCD (78, 90) = GCD (64, 567) = GCD (52, 790) = Least common multiple Def. Let a and b is the smallest positive integer that is divisible by both a and b, denoted by LCM (a, b). Example: Find the LCM of 120 = 2 * 3 * 5 and 500 = 2 * 5 2 3 LCM (120,500) = 2 max (3,* 3 max (1, * 5max (1, 3) = 2 * 3 * 5 = 20 LCM (40, 24) = LCM (147, 113) = LCM (145, 36) = LCM (74, 87) = LCM (177, 97) = LCM (186, 76) = LCM (153, 146) = LCM (52, 199) = LCM (279, 63) = LCM (265, 590) = Chapter 5.1 math induction How to do mathematical induction Step 1: Identify the math statement to be proven ????(????+1) Example: f(x) = 2 , where n is all real numbers Step 2: Show that the statement is true for the natural number 1 (or some smallest number) ????(????+1) Prove that f(x) = 2 will equal 1 Step 3: Show that if we assume that the statement is true for some k, then it follows that the statement must also be true for k+1 - Substitute all “n” for “k” ????(????+1) f(x) = 2 - Then substitute all “k” for “k+1” (????+1)((????+1)+1) f(x) = 2 Step 4: Conclude the proof by math induction… Example 1: Proof: ???? ???? ????+1 ) Step 1: Let P(n) be the statement ∑ ????=1 ???? = 2 This is the equation you need to solve Step 2: (Hint: write the basis step) Step 3: (Hint: assume and substitute) 1 ???? ???? Example 2: Prove that for all ???? ∈ ????,(1 + ) ≥ ???? + 2 ???? Proof: Let P(n)___________________________________________________ 1 1 1 Since (1 +2) = 1 + , 2(1) is true ( this is the basis step) Assume________________________. This means that ______________________________________. 1 ????+1 (1 + )2 =_______________________________ _______________________ 1 ???? ≥ (1 + )21 + ) 2 ________________________ =_______________________________ ________________________ =________________________________ ________________________ ????+1 ≥ 1 + 2 ____________________ Thus if P(k) is true, P(k+1) is also true. Hence by mathematical induction, ______________________________________________ Q.E.D. Chapter 6.1 Basics of counting Counting the complicated way … There are infinitely many ways to calculate the amount of probabilities and combinations Product rule n 1 n *2n *…3* n = n kwherexx is the product of all combinations Sum rule |A 1 A |2= |A | 1 |A | +2.+ |A | k Inclusive exclusion rule |A U B| = |A| + |B| - |A ∩ B| Using a tree diagram Example: How many different bit strings are there of length of seven? (Hint: use product rule) How many bit strings of length 10 begin and end with a 1? An office building contains 27 floors and has 37 offices on each floor. How many offices are in the building? How many different three-letter initials can people have? How many bit strings are there in length of six or less? How many strings of five ASCII characters contain the character “@” at least once (Hint, there is 128 ASCII characters) 6.2 Pigeonhole Principle Def. Imagine that 3 pigeons need to be placed into 2 pigeonholes. Can it be done? The answer is yes, but there is one catch. The catch is that no matter how the pigeons are placed, one of the pigeonholes must contain more than one pigeon. Example: 1. For every 27 word sequence in the US constitution, at least two words will start will the same letter. There are 27 words or “pigeons” that can start with one of the 26 different English letters or “pigeonholes.” By the pigeonhole principle, two of the words must start with the same letter 2. Two or more people reading this will have the same birthday. There are 366 possible birthdays (including February 29 in a leap year) and this study guide has many more than 15 readers. Therefore two of you must share the same birthday. 3. If you pick five cards from a standard deck of 52 cards, then at least two will be of the same suit. Each of the five cards can belong to one of four suits. By the pigeonhole principle, two or more must belong to the same suit. 4. If you have 10 black socks and 10 white socks, and you are picking socks randomly, you will only need to pick three to find a matching pair. The three socks can be one of two colors. By the pigeonhole principle, at least two must be of the same color. Another way of seeing this is by thinking sock by sock. If the second sock matches the first, then we are done. Otherwise, pick the third sock. Now the first two socks already cover both color cases. The third sock must be one of those and form a matching pair. ”If you're afraid to fail, then you're probably going to fail.” -Kobe Bryant FinalExamStudyguide 6.3 Combinations and permutations Permutation – elements in a particular order P(x, y) Example: Consider a set of elements containing a, b, and c, then we have sets (a,b,c) , (a,c,b) , (b,a,c), (b,c,a) , (c,a,b) , (c,b,a) As you can see each element is a rearrangement of the elements a, b, and c. Formula: ???? ????,???? =) ????! Where n and k are distinct integers (????−???? ! Exercise: How many rearrangements are there of an n-element set? (How many one-to-one functions?) Find the permutations for the following set of numbers P (78, 4) P (82, 3) P (6, 3) P (6, 5) P (8, 1) P (8, 5) P (8, 8) P (10, 9) P (6, 9) P (7, 65) P (4, 3) P (24, 12) P (14, 7) Combination – the number of combinations where the order does not matter C(x, y), can also be addressed as x “choose” y. ????! Formula: ???? ????,???? = ) ????! ????−???? !) Exercises: i) How many subsets of the alphabet contain exactly 3 letters? ii) How many subsets of the alphabet contain exactly 5 letters? iii) How many subsets of the alphabet contain at most 5 letters? iv) How many subsets of the alphabet contain more than 5 letters? Now that you know both permutation and combinations…. More Exercises: 1) How many permutations of the letters “abcdefghij” are there? 2) How many permutations of the letters “abcdefghij” contain a. “cde”? b. “bad”? c. “cab”? d. “gif”? 3) In how many ways can two students be chosen form a class 18 if one of the m receives an A and the other receives a B? 4) How many ways are there to select 5 players from a 10-member tennis team to make a trip to another school? 5) Find the number of ways to arrange 5 teacups that are chosen from a set of 7 different tea tables. 6) How many ways are there to select a committee to develop a secret world domination organization if the committee is to consist of 5 financial world economy investors and 9 notorious hackers, if there are only 10 investors and 13 hackers? 6.4 Binomial theorem Terminology: The number ( ) is also called a binomial coefficient because they occur as coefficients in ???? the expansion of powers of binomial expressions such as(???? + ????) . ???? Formula: An example on how this theorem works: (3x – 2) = C 10x)0 10–0(–2) + C103x1 10–1(–2) + C 10x2 10–2(–2)2 + C1033) 10–3(–2) + C103x4 10–4(–2) + C 10x5 10–5(–2)5 + C1036) 10–6(–2) + C103x7 10–7(–2) + C 10x8 10–8(–2)8 + C1039) 10–9(–2) + C103x10 10–10(–2) 10 Note how the highlighted counter number counts up from zero to 10, with the factors on the ends of each term having the counter number, and the factor in the middle having the counter number subtracted from 10. This pattern is all you really need to know about the Binomial Theorem; this pattern is how it works. How to find the coefficient of a certain term If the coefficient is needed for a certain term, use the binomial theorem to solve the coefficient 6 4 Where k is the term that is needed, for example x y , k would be either 6 or 4, the answer would be the same. “n” is the value of the binomial “a” is the first variable “b” is the second variable Exercises: Find the coefficient of 5 8 13 x y (x+y) 4 6 10 x y (x+y) 7 9 16 x y (2x-y) x y (3x+6) 15 10 15 25 x y (7x-8) x y (9x-3y) 16

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.