Ch 9.6 Counting, Permutations, and Combinations SKILLS OBJECTIVES Apply the fundamental counting principle to solve counting problems. Apply permutations to solve counting problems. Apply combinations to solve counting problems. CONCEPTUAL OBJECTIVE Understand the difference between permutations and combinations. The Fundamental Counting Principle You are traveling through Europe for the summer and decide the best packing option is to select separates that can be mixed and matched. You pack one pair of shorts and one pair of khaki pants. You pack a pair of Teva sport sandals and a pair of sneakers. You have three shirts (red, blue, and white). How many different outfits can be worn using only the clothes mentioned above? The answer is 12. There are two options for bottoms (pants or shorts), three options for shirts, and two options for shoes. The product of these is 232 = 12 1
The general formula for counting possibilities is given by the fundamental counting principle. Fundamental Counting Principle definition: Let E 1 and E 2 be two independent events. The first event E 1 can occur in m 1 ways. The second event E 2 can occur in m 2 ways. The number of ways that the combination of the two events can occur is m1 m2. In other words, the number of ways in which successive things can occur is found by multiplying the number of ways each thing can occur. Study Tip: The fundamental counting principle can be extended to more than two events. EXAM PLE 1 Possible Meals Served at a Restaurant A restaurant is rented for a retirement party. The owner offers an appetizer, an entree, and a dessert for a set price. The following are the choices that people attending the party may choose from. How many possible dinners could be served that night? Appetizers: calamari, stuffed mushrooms, or caesar salad Entrées: tortellini alfredo, shrimp scampi, eggplant parmesan, or chicken marsala Desserts: tiramisu or flan Solution: Strategy: Tactics and Calculations: Write in English Language, the Answer: There are three possible appetizers, four possible entrées, and two possible desserts. Write the product of possible options. 3 4 2 = 24 There are 24 possible dinners for the retirement party 2
YOUR TURN In Example 1, the restaurant will lower the cost per person for the retirement party if the number of appetizers and entrées is reduced. Suppose the appetizers are reduced to either soup or salad and the entrées are reduced to either tortellini or eggplant parmesan. How many possible dinners could be served at the party? Original Problem Information Appetizers: calamari, stuffed mushrooms, or caesar salad Entrées: tortellini alfredo, shrimp scampi, eggplant parmesan, or chicken marsala Desserts: tiramisu or flan Solution: Strategy: Tactics and Calculations: Write in English Language, the Answer: There are two and not three possible appetizers, two and not four possible entrées, and two possible desserts. Write the product of possible options. Extract from given English lanuage information the mathematical information. 2 appetizers 2 entrees 2 deserts 2 2 2 = 8 There are 24 possible dinners for the retirement party 3
EXAMPLE 2 Telephone Numbers (When to Require 10-Digit Dialing) In many towns in the United States, residents can call one another using a 7-digit dialing system. In some large cities, 10-digit dialing is required because two or more area codes coexist. Determine how many telephone numbers can be allocated in a 7-digit dialing area. Solution: Strategy: Tactics and Calculations: With 7-digit telephone numbers, the first number cannot be a 0 or a 1, but each of the following six numbers can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. First number: 2, 3, 4, 5, 6, 7, 8, or 9. 8 possible digits Second number: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. 10 possible digits Third number: 10 possible digits Fourth number: 10 possible digits Fifth number: 10 possible digits Sixth number: 10 possible digits Seventh number: 10 possible digits Counting principle: 8 10 10 10 10 10 10 Formula and Calculations Numerical answer. Possible telephone numbers: 8,000,000 Write in English Eight million 7-digit telephone numbers can be language the answer: allocated within one area code. 4
YOUR TURN If the first digit of an area code cannot be 0 or 1, but the second and third numbers of an area code can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9, how many 10-digit telephone numbers can be allocated in the United States? Answer: 6.4 billion 5
The fundamental counting principle applies when an event can occur more than once. We now introduce two other concepts, permutations and combinations, which allow individual events to occur only once. For example, in Example 2, the allowable telephone numbers can include the same number in two or more digit places, as in 555-1212. However, in many state lottery games, once a number is selected, it cannot be used again. An important distinction between a permutation and a combination is that in a permutation order matters, but in a combination order does not matter. For example, the Florida winning lotto numbers one week could be 2-3-5-11-19-27. This would be a combination because the order in which they are drawn does not matter. However, if you were betting on a trifecta at the Kentucky Derby, to win you must not only select the first, second, and third place horses, you must select them in the order in which they finished. This would be a permutation. 6
Permutations DEFINITION Permutation A permutation is an ordered arrangement of distinct objects without repetition. EXAMPLE 3 Finding the Number of Permutations of n Objects How many permutations are possible for the letters A, B, C, and D? Solution: ABCD ABDC ACBD ACDB ADCB ADBC BACD BADC BCAD BCDA BDCA BDAC CABD CADB CBAD CBDA CDAB CDBA DABC DACB DBCA DBAC DCAB DCBA There are 24 (or 4!) possible permutations of the letters A, B, C, and D. Notice that in the first row of permutations in Example 3, A was selected for the first space. That left one of the remaining three letters to fill the second space. Once that was selected there remained two letters to choose from for the third space, and then the last space was filled with the unselected letter. In general, there are n! ways to order n objects. 7
Number of Permutations of n Objects The number of permutations of n objects is n! = n n 1 n 2 21 ( ) ( ) EXAMPLE 4 Running Order of Dogs In an American Kennel Club (AKC) sponsored field trial, the dogs compete in random order. If there are nine dogs competing in the trials, how many possible running orders are there? Solution: Strategy: There are nine dogs that will run. It does not matter which order they run in (random per given information.) Tactics and Formula: The number of possible running orders is n! because after first dog runs then there are 8 dogs, then 7 and so forth. Calculations: Write in English language the answer. Number of dogs competing is 9 so let n=9. n!= ; write formula 9! ; substitute into formula 9 for n 9! = 9 8 7 6 5 4 3 2 1 = 362,880 There are 362,880 different possible running orders of nine dogs. 8
YOUR TURN Five contestants in the Miss America pageant reach the live television interview round. In how many possible orders can the contestants compete in the interview round? Solution: C.Root in Common Core or IB like format: Strategy: There are five contestants that will participate in the interview round. The number of possible order of choosing a contestant is being asked. We know that once a contestant is asked they will not be asked again. So a factorial problem. We also know that this fits the description of a permutation which by definition is an ordered arrangement of distinct objects without repetition. Tactics and Formula: The number of possible running orders is n! to be a factorial problem. Because after first contestant interviews then there are 4, then 3, then 2 and finally 1. Calculations: Number of contestants competing is 5 so let n=5. n!= ; write formula 5! ; substitute into formula 9 for n 5! = 5 4 3 2 1 = 120 Write in English There are 120 different possible interviewing language the answer. orders of five contestants. 9
Number of Permutations of n Objects Taken r at a Time The number of permutations of n objects taken r at a time is n! np = r n( n 1)( n 2) ( n r 1) n r! = + ( ) EXAMPLE 5 Starting Lineup for a Vollyball Team The starters for a six-woman volleyball team have to be listed in a particular order (1-6). If there are 13 women on the team, how many possible starting lineups are there? 10
YOUR TURN A softball team has 12 players, 10 of whom will be in the starting lineup (batters 1-10). How many possible starting lineups are there for this team? Answer: 239,500,800 Combinations The difference between a permutation and a combination is that a permutation has an order associated with it, whereas a combination does not have an order associated with it. DEFINITION Combination A combination is an arrangement, without specific order, of distinct objects without repetition. The six winning Florida lotto numbers and the NCAA men s Final Four basketball tournament are examples of combinations (six numbers and four teams) without regard to order. The number of combinations of n objects taken r at a time is equal to the binomial n coefficient r. Number of Combinations of n Objects Taken r at a Time The number of combinations of n objects taken r at a time is n n! ncr = = r ( n r)! r! 11
Compare the number of permutations n! the number of combinations n r ( ) C np = r n( n 1)( n 2) ( n r 1) ( n r)! = + n n! = = r n r! r!. and It makes sense that the number of combinations is less than the number of permutations. The denominator is larger because there are no separate orders associated with a combination. Lesson for greater understanding by C.Root: In other words, in a combination, the denominator value reduces the numerator value by r!. If Order of Selection is Important, then a Permutation. If Order of Selection is Unimportant, then a Combination. EXAMPLE 3 Finding the Number of Permutations of n Objects How many permutations are possible for the letters A, B, C, and D? Solution: ABCD ABDC ACBD ACDB ADCB ADBC BACD BADC BCAD BCDA BDCA BDAC CABD CADB CBAD CBDA CDAB CDBA DABC DACB DBCA DBAC DCAB DCBA There are 24 (or 4!) possible permutations of the letters A, B, C, and D. Let s place this in to the permutation equation: n! 4! 4321 np = r 24 permutations ( n r)! = ( 4 4 )! = 0! = And now let s place this data into a combination equation where selection does not matter. n n! 4! 4! 1 4! ncr = = = = = = 1 ; 1 combination r ( n r)! r! ( 4 4 )!4! 0!4! 1 4! 12
In English language answer: Thus the number of permutations where every selection is important results in 24 permutations, but with a combination, where a selection does not matter, then only 1combination is necessary. Textbook EXAMPLE 6 Possible Combinations to the Lottery If there are a possible 59 numbers and the lottery officials draw 6 numbers, how many possible combinations are there? 13
YOUR TURN What are the possible combinations fora lottery with 49 possible numbers and 6 drawn numbers? 14
Permutations with Repetition Permutations and combinations are arrangements of distinct (nonrepeated) objects. A permutation in which some of the objects are repeated is called a permutation with repetition or a nondistinguishable permutation. Number of Distinguishable Permutations definition If a set of n objects has n 1 of one kind of object, n 2 of another kind of object, n 3 of a third kind of object, and so on for k different types of objects so that n= n1+ n2 + + nk, then the number of distinguishable permutations of the n objects is n! n! n! n! n! 1 2 3 k 15
For example, if a sack has three red marbles, two blue marbles, and one white marble, how many possible permutations would there be when drawing six marbles, one at a time? This is a different problem from writing the numbers 1 through 6 on pieces of paper putting them in a hat, and drawing them out. The reason the problems are different is that the two blue balls are indistinguishable and the three red balls are also indistinguishable. The possible permutations for drawing numbers out of the hat are 6!, whereas the possible permutations for drawing balls out of the sack are given by 6! 3! 2! 1! In our sack of marbles, there were six marbles n = 6. Specifically, there were three red marbles (n1 = 3) two blue marbles (n2 = 2) and one white marble (n3 = 1). Notice that n = n1 + n2 + n3 and that the number of distinguishable permutations is equal to Let s put this example in relation to the formula by first writing the formula and then substituting values. n! ; Formula for Number of Distinguisable Permutations n! n! n! n! 1 2 3 k ( n1 = 3 red marbles + n2 = 2 blue marbles + n3 = 1 white marbles )! ( n = ) ( n = ) ( n = ) 3 red marbles! 2 blue marbles! 1 white marble! 1 2 3 654321 720 = = = 60 3 2 1 2 1 1 12 = 16
EXAMPLE 7 Peg Game at Cracker Barrel The peg game on the tables at Cracker Barrel is a triangle with 15 holes drilled in it, in which pegs are placed. There are 5 red pegs, 5 white pegs, 3 blue pegs, and 2 yellow pegs. If all 15 pegs are in the holes, how many different ways can the pegs be aligned? Solution: There are four different colors of pegs (red, white, blue, and yellow). There are a possible 567, 560 ways to insert the 15 colored pegs at the Cracker Barrel. 17
YOUR TURN Suppose a similar game to the peg game at Cracker Barrel is set up with only ten holes in a triangle. With 5 red pegs, 2 white pegs, and 3 blue pegs, how many different permutations can fill that board? Answer; 2520 18
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