Report

Chapter 10 Ways to Count- Tree Diagram A sporting goods store offers 3 types of snowboards (all-mountain, freestyle, and carving) and 2 types of boots (soft and hybrid). How many choices does the store offer for snowboarding equipment? A sporting goods store offers 3 types of snowboards (all-mountain, freestyle, and carving) and 2 types of boots (soft and hybrid). How many choices does the store offer for snowboarding equipment? Soft All-mountain board, soft boots Hybrid all-mountain board, hybrid boots Soft Freestyle board, soft boots Hybrid Freestyle board, hybrid boots Soft Carving board, soft boots Hard Carving board, hard boots All-mountain Freestyle Carving Ways to count- Tree Diagram The tree has 6 branches. So, there are 6 possible choices! Use a Tree Diagram The snowboard shop also offers 3 different types of bicycles (mountain, racing, and BMX) and 3 different wheel sizes (20”, 22”, and 24”). How many bicycle choices does the store offer? ANSWER 9 Bicycles Fundamental Counting Principle Snowboarding example There are 3 options for boards and 2 options for boots. So the number of possibilities is 32=6 This is called the “Fundamental Counting Principle” Fundamental Counting Principle Two events If one event can occur in m ways and another event can occur in n ways, then the number of ways that both events can occur is mn. Three or more events The fundamental counting principle extends to 3 or more events. The product of possibilities for each event will give the possibility of ALL the events occurring. Use the Fundamental Counting Principle You are framing a picture. The frames are available in 12 different styles. Each style is available in 55 different colors. You also want a blue mat board, which is available in 11 different shades of blue. How many different ways can you frame the picture? Use the Fundamental Counting Principle Multiply # of frame styles (12) # of frame colors (55) # of mat boards (11) 12 x 55 x 11 = 7260 There are 7260 different ways to frame the picture! Use the Counting Principle with or without Repetition The standard configuration for a Texas license plate is 1 letter followed by 2 digits followed by 3 letters Use the Counting Principle with Repetition How many different license plates are possible if letters and digits can be repeated? Use the Counting Principle with Repetition How many different license plates are possible if letters and digits can be repeated? Multiply # of letters possible (26) # of digits possible (10) # of digits possible (10) # of letters possible (26) # of letters possible (26) # of letters possible (26) Use the Counting Principle with Repetition 26 x 10 x 10 x 26 x 26 x 26 = 45,697,600 Use the Counting Principle without Repetition How many different license plates are possible if letters and digits cannot be repeated? Use the Counting Principle without Repetition How many different license plates are possible if letters and digits cannot be repeated? Multiply # of letters possible (26) # of digits possible (10) # of digits possible w/o the previous digit used (9) # of letters possible w/o the first letter used (25) # of letters possible w/o the first 2 letters used (24) # of letters possible w/o the first 3 letters used (23) Use the Counting Principle without Repetition 26 x 10 x 9 x 25 x 24 x 23 = 32,292,000 More Counting Principles with and without Repetition How would the answers change for the standard configuration of a New York license plate, which is 3 letters followed by 4 numbers? How many possibilities with repetition? How many possibilities without repetition? Answer With Repetition- 175,760,000 possibilities Without Repetition- 78,624,000 possibilities Permutations Question- How many ways can you rearrange the word COW COW CWO OCW OWC WOC WCO Permutations The answer is 6 times. The ordering of a certain number of n objects is called a PERMUTATION. Permutations Back to the question… “How many ways can you rearrange the word COW?” You can use the fundamental counting principle to find the number of permutations of C, O, and W. First letter- 3 choices Second letter- 2 choices Third letter- 1 choice Permutations Therefore, we find the number of permutations is 3 x 2 x 1 = 6 Permutations How many permutations for the word: SEA ROCK MATH Factorials! 3 x 2 x 1 can also be written as 3! ! is the symbol used for factorial 3! is read “three factorial” Factorials! In general, n! is defined where n is a positive integer as follows: n! = n (n-1) (n-2) … 3 2 1 Factorials! 5! = 5 4 3 2 1 = 120 12! = 12 11 10 9 8 7 6 5 4 3 2 1 = 479,001,600 Permutations using Factorials Ten teams are competing in the final round of the Olympic four-person bobsledding competition. In how many different ways can the bobsledding teams finish the competition? (Assume no ties) Permutations with Factorials There are 10! ways that teams can finish. 10! = 10 9 8 7 6 5 4 3 2 1 = 3,628,800 Permutations with Factorials How many different ways can 3 of the bobsledding teams finish first, second, and third to win the gold, silver, and bronze medals? Permutations with Factorials Any of the 10 teams can finish first, then any of the remaining 9 teams can finish second, and finally any of the 8 remaining teams can finish third. So number of ways teams can win medals is 10 9 8 = 720 Permutations with Factorials How would the answers change if there were 12 bobsledding teams competing in the final round of he competition? A- How many was can the teams finish? B- How many ways to win medals? Answer A- 12! = 479,001,600 B- 12 11 10 = 1320 Permutations of n Objects Taken r at a Time The answer to the second part of the bobsledding questions is called the number of permutations of 10 objects taken 3 at a time. Denoted as 10P3 10P3 = 10 9 8 = 10 ·9 ·8· 7·6·5· 4·3·2 ·1 7·6 ·5· 4·3·2·1 10! 10! = = 7! (10 - 3)! Permutations with Repetition The general equation for the number of permutations of n objects taken r at a time is: n! n Pr = (n - r)! Use the Permutation Equation You a burning a demo CD for your band. Your band has 12 songs stored on your computer. However, you want to put only 4 songs on the demo CD. In how many orders can you burn 4 of the 12 songs onto the CD? Use the Permutation Equation 12! 12! 479, 001, 600 = = =11,880 12 P4 = (12 - 4)! 8! 40,320 Try: 5P3 4P1 8P5 Practice Answers 5P 3 = 60 4P 1 = 4 8P 5 = 6720 Permutations with Repetition How many ways can you write the word MOO Note: Each “O” is a distinct letter. Permutations with Repetition MOO MOO OMO OMO OOM OOM There are 6 permutations of the letters M, O, and O. However, if the two occurrences of O are considered interchangeable, then there are only three distinguishable permutations: MOO OMO OOM Permutations with Repetition Each of these permutations corresponds to two of the original six permutations because there are 2! , or 2, permutations of O and O. So the number of permutations of M, O, and O can be written as 3! = 6 = 3 2! 2 Permutations with Repetition To generalize the situation we just explored we can say that the number of distinguishable permutations of n objects where one object is repeated s1 times, another is repeated s2 times, and so on, is: n! s1 !· s2 !·...· sk ! Permutations with Repetition How many ways can you rearrange the letters of the following words: MATHEMATICS ALGEBRA MISSISSIPPI TALLAHASSEE Permutations with Repetition MATHEMATICS 11! 39, 916,800 = = 4, 989, 600 2!·2!·2! 2 ·2 ·2 ALGEBRA 7! 5, 040 = = 2, 520 2! 2 Permutations with Repetition MISSISSIPPI 11! 39, 916,800 = = 34, 650 4!· 4!·2! 24·24·2 TALLAHASSEE 11! 39, 916,800 = = 831, 600 3!·2!·2!·2! 6·2·2 ·2 Permutations vs Combinations With permutations, we learned that order is important for some counting problems. For other counting problems, order is not important. For example, if you purchase a package of trading cards, the order of the cards inside the package is not important. Permutations vs. Combinations COMBINATION- a selection of r objects from a group of n objects where order is NOT important. Think of a combination pizza; you have various kinds of toppings (pepperoni, olives, green peppers, etc.) and you can put them on in several different orders, but in the end, order doesn’t matter. Its just one delicious pizza! Combination Equation The number of combinations of r objects taken from a group of n distinct objects is denoted by nCr and is given by this formula: n! n Cr = (n - r)!·r! nC r is read “n choose r” Find Combinations A standard deck of 52 playing cards has 4 suits with 13 different cards in each suit. SUIT- hearts, clubs, spades, and diamonds Find Combinations Each suit has the following cards called “kinds”: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King There is one of each of these kinds in each suit Face Cards- Cards that have faces on them (Jack, Queen, King) Find Combinations If the order in which the cards are dealt is not important, how many different 5card hands are possible? Remember… n! n Cr = (n - r)!·r! In this example, n= 52 cards and r = 5 cards Find Combinations Therefore, 52 C5 = 52! 52! 52 ·51·50· 49· 48· 47! = = = 2, 598, 960 (52 - 5)!·5! 47!·5! 47!·5! Find Combinations In how many 5-card hands are all 5 cards of the same color? Find Combinations For all 5 cards to be the same color, you need to choose 1 of the 2 colors and then 5 of the 26 cards in that color. So, the number of possible hands is: 2! 26! 2 26 ·25·24·23·22 ·21! · = · =131, 560 2 C 1· 26 C5 = 1!·1! 21!·5! 1·1 21!·5! Multiple Events When finding the number of ways both an event A and an event B can occur, you need to multiply. When finding the number of ways that event A or event B can occur, you add instead. Add or Multiply Combinations William Shakespeare wrote 38 plays that can be divided into 3 genres. Of the 38 plays, 18 are comedies, 10 are histories, and 10 are tragedies. How many different sets of exactly 2 comedies and 1 tragedy can you read? How many different sets of at most 3 plays can you read? Add or Multiply Combination How many different sets of exactly 2 comedies and 1 tragedy can you read? You can choose 2 of the 18 comedies and 1 of the 10 tragedies. So, the number of possible sets of plays is: 18 C 2 ·10 C1 = 18! 10! 18·17·16! 10·9! · = · =153·10 =1, 530 16!·2! 9!·1! 16!·2·1 9!·1 Add or Multiply Combinations How many different sets of at most 3 plays can you read? You could read 0, 1, 2, or 3 plays. Because there are 38 plays that can be chosen, the number of possible sets of plays is: 38 C0 + 38 C1 + 38 C2 + 38 C3 =1+38+ 703+8, 436 = 9,178 Practice How many different sets of exactly 3 tragedies and 2 histories can you read? 8C 3 10C6 7C 2 Practice Answers 5400 sets 8C 3 = 56 10C6 = 7C 2 = 210 21 Subtracting Possibilities Counting problems that involve phrases like “at most” or “at least” are sometimes easier to solve by subtracting possibilities you do not want from the total number of possibilities. The portion NOT included in the number of possibilities is also called the COMPLIMENT. Subtracting Possibilities During the school year, the girl’s basketball team is scheduled to play 12 home games. You want to attend at least 3 of the games. How many different combinations of games can you attend? Subtracting Possibilities Of the 12 home games, you want to attend 3 games, or 4 games, or 5 games, and so on. So, the number of combinations of games you can attend is: C + C + C +... + C 12 3 12 4 12 5 12 12 Subtracting Possibilities Instead of adding these combinations, use the following reasoning. For each of the 12 games, you can choose to attend or not attend the game, so there are 212 total combinations. If you attend at least 3 games, you do not attend only a total of 0, 1, or 2 games. So, the number of ways you can attend at least 3 games is: 2 - ( 12 C0 + 12 C1 + 12 C2 ) = 4, 096 - (1+12 + 66) = 4, 017 12 Pascal’s Triangle If you arrange the values of nCr in a triangular pattern in which each row corresponds to a value of n, you get what is called Pascal’s triangle, named after the French mathematician Blaise Pascal. Pascal’s Triangle Pascal’s triangle is show below with its entries represented by combinations and with its entries represented by numbers The first and last numbers in each row are 1. Every number other than 1 is the sum of the closest two numbers in the row directly above it. Pascal’s Triangle as numbers Pascal’s Triangle as combinations Use Pascal’s Triangle The 6 members of a Model UN club must choose 2 representatives to attend a state convention. Use Pascal’s triangle to find the number of combinations of 2 members that can be chosen as representatives. Use Pascal’s Triangle Because you need to find 6C2, write the 6th row of Pascal’s triangle by adding numbers from the previous row. n = 5 (5th row) 1 5 10 10 5 1 n = 6 (6th row) 1 6 15 20 15 6 1 The value of 6C2 is the 3rd number in the 6th row of Pascal’s triangle, as show above. Therefore, 6C2 = 15. There are 15 combinations of representatives for the convention. Binomial Expansions There is an important relationship between powers of binomials and combinations. The numbers in Pascal’s triangle can be used to find coefficients in binomial expansions. Binomial Expansions For example, the coefficients in the expansion (a + b)4 are the numbers of combinations in the row of Pascal’s triangle for n = 4: (a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4 4C 0 4C1 4C2 4C3 4C4 This result is generalized in the binomial theorem. Binomial Theorem For any positive integer n, the binomial expansion of (a + b)n is: (a + b)n = nC0anb0 + nC1an-1b1 + nC2an-2b2 +…+ nCna0bn Notice that each term in the expansion of (a + b)n has the form nCran-rbr where r is an integer from 0 to n. Expand a power of a binomial sum Use the binomial theorem to write the binomial expansion for: (x2 + y)3 Expand a power of a binomial sum (x 2 + y)3 = 3 C0 (x 2 )3 y0 + 3 C1(x 2 )2 y1 + 3 C2 (x 2 )1 y2 + 3 C3 (x 2 )0 y3 = (1)(x 6 )(1)+ (3)(x 4 )(y)+ (3)(x 2 )(y2 )+ (1)(1)(y3 ) = x + 3x y + 3x y + y 6 4 2 2 3 Expand a power of a binomial difference Use the binomial theorem to write the binomial expansion of: (a-2b)4 Expand a power of a binomial difference (a - 2b)4 = [a + (-2b)]4 = 4 C0 a4 (-2b)0 + 4 C1a3 (-2b)1 + 4 C2 a2 (-2b)2 + 4 C3a1 (-2b)3 + 4 C4 a0 (-2b)4 = (1)(a4 )(1)+(4)(a3 )(-2b)+(6)(a2 )(4b2 )+(4)(a)(-8b3 )+(1)(1)(16b4 ) = a -8a b + 24ab - 32ab +16b 4 3 2 3 4 Powers of Binomial Differences To expand a power of a binomial difference, you can rewrite the binomial as a sum. The resulting expansion will have terms whose signs alternate between + and -. Use the binomial theorem to write the binomial expansion (x + 3)5 (a + 2b)4 (2p – q)4 (5 – 2y)3 (x + 3)5 = 5 C0 a5 (3)0 + 5 C1a4 (3)1 + 5 C2 a3 (3)2 + 5 C3a2 (3)3 + 5 C4 a1 (3)4 + 5 C5a0 (3)5 = (1)a 5 (1) + (5)a 4 (3) + (10)a3 (9) + (10)a 2 (27) + (5)a(81) + (1)(1)(243) = a 5 +15a 4 + 90a3 + 270a 2 + 405a + 243 (a + 2b)4 = 4 C0 a4 (2b)0 + 4 C1a3 (2b)1 + 4 C2 a2 (2b)2 + 4 C3a1 (2b)3 + 4 C4 a0 (2b)4 = (1)(a4 )(1)+(4)(a3 )(2b)+(6)(a2 )(4b2 )+(4)(a)(8b3 )+(1)(1)(16b4 ) = a +8a b + 24ab + 32ab +16b 4 3 2 3 4 Find a coefficient in an expansion Find the coefficient of x4 in the expansion of (3x + 2)10 Find a coefficient in an expansion From the binomial theorem, you know the following: (3x + 2) = 10 C0 (3x) (2) + 10 C1 (3x) (2) +... + 10 C10 (3x) (2) 10 10 0 9 1 Each term in the expansion has the form 10-r r 4 C (3x) (2) The term containing x 10 r occurs when r = 6: 4 6 4 4 C (3x) (2) = (210)(81x )(64) =1, 088, 640x 10 6 0 10 Find the coefficient in an expansion So the coefficient of x4 is 1,088,640 Find the coefficient of: x5 in the expansion of (x – 3)7 x3 in the expansion of (2x + 5)8 Practice Answers x5 in the expansion of (x – 3)7 189 x3 in the expansion of (2x + 5)8 1,400,000