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Permutations and Combinations Review Plus a little Dry Run, Sorting and Code.

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Presentation on theme: "Permutations and Combinations Review Plus a little Dry Run, Sorting and Code."— Presentation transcript:

1 Permutations and Combinations Review Plus a little Dry Run, Sorting and Code.

2 Objectives:  Review Permutations and Combinations  apply fundamental counting principle  compute permutations  compute combinations  distinguish permutations vs combinations  Practice reading methods that use parameters  Review Insertion Sort  Write a program that incorporates the insertion sort.

3 Fundamental Counting Principle Fundamental Counting Principle can be used determine the number of possible outcomes when there are two or more characteristics. Fundamental Counting Principle states that if an event has m possible outcomes and another independent event has n possible outcomes, then there are m * n possible outcomes for the two events together.

4 Fundamental Counting Principle Lets start with a simple example. A student is to roll a die and flip a coin. How many possible outcomes will there be? 1H 2H 3H 4H 5H 6H 1T 2T 3T 4T 5T 6T 12 outcomes 6*2 = 12 outcomes

5 Fundamental Counting Principle For a college interview, Robert has to choose what to wear from the following: 4 slacks, 3 shirts, 2 shoes and 5 ties. How many possible outfits does he have to choose from? 4*3*2*5 = 120 outfits

6 Permutations A Permutation is an arrangement of items in a particular order. Notice, ORDER MATTERS! To find the number of Permutations of n items, we can use the Fundamental Counting Principle or factorial notation.

7 Permutations The number of ways to arrange the letters ABC: ____ ____ ____ Number of choices for first blank ? 3 ____ ____ 3 2 ___ Number of choices for second blank? Number of choices for third blank? 3 2 1 3*2*1 = 6 3! = 3*2*1 = 6 ABC ACB BAC BCA CAB CBA

8 Permutations To find the number of Permutations of n items chosen r at a time, you can use the formula

9 Permutations A combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated? Practice: Answer Now

10 Permutations A combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated? Practice:

11 Permutations From a club of 24 members, a President, Vice President, Secretary, Treasurer and Historian are to be elected. In how many ways can the offices be filled? Practice: Answer Now

12 Permutations From a club of 24 members, a President, Vice President, Secretary, Treasurer and Historian are to be elected. In how many ways can the offices be filled? Practice:

13 Combinations A Combination is an arrangement of items in which order does not matter. ORDER DOES NOT MATTER! Since the order does not matter in combinations, there are fewer combinations than permutations. The combinations are a "subset" of the permutations.

14 Combinations To find the number of Combinations of n items chosen r at a time, you can use the formula

15 Combinations To find the number of Combinations of n items chosen r at a time, you can use the formula

16 Combinations To play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible? Practice: Answer Now

17 Combinations To play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible? Practice:

18 Combinations A student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions? Practice: Answer Now

19 Combinations A student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions? Practice:

20 Combinations A basketball team consists of two centers, five forwards, and four guards. In how many ways can the coach select a starting line up of one center, two forwards, and two guards? Practice: Answer Now

21 Combinations A basketball team consists of two centers, five forwards, and four guards. In how many ways can the coach select a starting line up of one center, two forwards, and two guards? Practice: Center: Forwards:Guards: Thus, the number of ways to select the starting line up is 2*10*6 = 120.

22 Summary Permutations: Order matters The number of Permutations of n items chosen r at a time, you can use the formula Combinations: Order does not matter. To find the number of Combinations of n items chosen r at a time, you can use the formula

23 Stats Pack: Write a program with methods to calculate the following. Factorial –Sent a positive integer, returns its factorial. (Note: 0! = 1) Combinations Write a method that will find the number of combinations of n items taken r at a time. Combinations of n items take r at a time= n!/((r!)(n-r)!) –Enter the values for n and r in the main body –Calculate and return the number of combinations in the method –Show the number of combinations in the main body. Permutations Write a method that will find the number of permutations of n items taken r at a time. Permutations of n items taken r at a time= n!/(n-r)! –Enter the values for n and r in the main body –Calculate and return the number of combinations in the method –Show the number of combinations in the main body.

24 Learning Objectives Be able to read a methods that are passed parameters. Be able to write a second program that uses methods.

25 Multiple Choice CTA9 (none). Consider the following static method public static double getSomething(int val) { val = 2 * val; val = val + 2 * val; return val; } Which of the following could be used to replace the body of getSomething so that the modified version will return the same result as the original version for all values of the parameter val. A.return 2 * val; B.return 3 * val; C.return 4 * val; D.return 5 * val; E.return 6 * val;

26 Multiple Choice CTA10 (A1.15). Consider the following static method public static double getSomething(int val) { val = 2 + val; val = val + 3*val; return val; } Which of the following could be used to replace the body of getSomething so that the modified version will return the same result as the original version for all values of the parameter val. A.return 4*val + 2; B.return 4*val + 6; C.return 4*val + 8; D.return 7*val + 6; E.return 7*val + 8;

27 public static void sort(String [] list) { for(int start = 0; start < list.length-1; start++) { int index = start; for(int k = start + 1; k < list.length; k++) { if(list[k].compareTo(list[index]) > 0) index = k; } String temp = list[start]; list[start] = list[index]; list[index] = temp; } Assume the String array words is initialized as shown below. word What will the array word look like after the third pass through the outer loop in the call sort(word)? Note how an array is passed to a method. BillJoeSueAnnMelZeb

28 Multiple Choice A20 (A3.32). Consider the incomplete method ChoiceOfThree. public int choiceOfThree() { // missing code } Method choiceOfThree is intended to return one of the values 1, 2, 3, each with probability 1/3. Consider the following replacements for //missing code. I.return (int)(Math.random() * 3); II.return (int)(Math.random() * 3) + 1; III.if(Math.random() < 1.0/3.0) return 1; else if (Math.random() < 2.0/3.0) return 2; else return 3; Which if these replacements for //missing code will make choiceOfThree work as intended? A. I onlyB. II onlyC. III onlyD. I and IIIE. II and III

29 Insertion sort How it works Example LowHigh 8 6 7 3 15 1 5 Stability Speed Your turn HighLow 8 2 5 3 9 4 6 1 7 Dummy, slide, back Stable O(n 2 )

30 Insertion Sort // a is the name of the array // nElems stores the number of elements being sorted // This example is for sorting an array of ints int in, out; for(out=1; out<nElems; out++) // out is dividing line { int dummy = a[out]; // dummy Need to modify for sorting different types in = out; // start shifts at out while(in>0 && a[in-1] >= dummy) // until one is smaller, What if sorting Strings? { a[in] = a[in-1]; // Slide: shift item right, in--; // go left one position } a[in] = dummy; // Back: insert marked item } // end for

31 Second Method Program Main Body –Get 9 scores –Call methods –Show results Scores in order Low to High Mean, median Push: Mode, standard deviation Push: Handle getting an unknown number (but less than 100) of values. Methods –Insertion Sort –Mean –Median –Push: Mode, standard deviation


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