COUNTING TECHNIQUES PERMUTATIONS AND COMBINATIONS Computer Science Statistics
COUNTING TECHNIQUES PERMUTATIONS AND COMBINATIONS
Computer Science, Statistics and Probability all involve counting techniques which are a branch of mathematics called combinatorics (ways to combine things). We'll be introducing this topic in this section. For dinner you have the following choices: ENTREES MAINS soup salad DESSERTS icecream chicken prawns hamburger How many different combinations of meals could you make? We'll build a tree diagram to show all of the choices.
Notice the number of choices at each branch 2 choice s 3 choice s 2 x 3 x 2 = 12 p u so sa 2 choice s en k c chi prawns ham bur ger ice cream la d n e k c chi prawns ham bur Now to get all possible choices we follow each path. ger ice cream We ended up with 12 possibilities soup, chicken, ice cream soup, chicken, soup, prawns, ice cream soup, prawns, soup, hamburger, ice cream soup, hamburger, salad, chicken, ice cream salad, chicken, salad, prawns, ice cream salad, prawns, salad, hamburger, ice cream salad, hamburger,
Multiplication Principle of Counting If a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in px qx r different ways. If we have 6 different shirts, 4 different pants, 5 different pairs of socks and 3 different pairs of shoes, how many different outfits could we wear? 6 x 4 x 5 x 3= 360 Much quicker than making a list!
Choosing a path……how many ways from A to C through B? A to B = 4 ways B to C = 2 ways 4 x 2 = 8 ways
You try! 5 X 4 X 3 = 60 ways
A little tougher…. .
A permutation is an ordered arrangement of r objects chosen from n objects. For combinations order does not matter but for permutations it does. There are three types of permutations. The first is distinct with repetition. This means there are n distinct objects but in this means different choosing r of them you can repeat an object. Let's look at a 3 There are 10 choices for the first number combination lock There are 10 choices for the second number with numbers 0 and you can repeat the first number through 9 There are 10 choices for the third number and you can repeat By the multiplication principle there are 10 x 10 = 1000
This can be generalized as: Permutations: Distinct Objects with Repetition The number of ordered arrangements of r objects chosen from n objects, in which the n objects are distinct and repetition is allowed, is nr What if the lock had four choices for numbers instead of three? 104 = 10 000 choices
The second type of permutation is distinct, without repetition. Let's say four people have a race. Let's look at the possibilities of how they could place. Once a person has been listed in a place, you can't use that person again (no repetition). Based on the multiplication principle: First place would be choosing someone from 4 x 3 x 2 x 1 = 24 choices among 4 people. Now there are only 3 to choose from for second place. Now there are only 2 to choose from for third place. Only one possibility for fourth place. 4 th 3 rd 2 nd 1 st
n. P , means the number of ordered arrangements of r objects chosen from n distinct objects and repetition is not allowed. r In the last example: If you have 10 people racing and only 1 st, 2 nd and 3 rd place how many possible outcomes are there? 0! = 1
A combination is an arrangement of r objects chosen from n objects regardless of order. n. C , means the number combinations of r objects chosen from n distinct objects and repetition is not allowed. r Order doesn't matter here so the combination 1, 2, 3 is not different than 3, 2, 1 because they both contain the same numbers. Note: Dividing out by the common “r” combinations. Hence, you will have fewer combinations than permutations!
You need 2 people on your committee and you have 5 to choose from. You can see that this is without repetition because you can only choose a person once, and order doesn’t matter. You need 2 committee members but it doesn't matter who is chosen first. How many combinations are there?
The third type of permutation is involving n objects that re are not distinct. ex pre How many different combinations of letters in specific am se n order (but not necessarily English words) can be formed ple tati ve using ALL the letters in the word REARRANGE? Not Examinable. . Just for Fun E R N R A A G E R The "words" we form will have 9 letters so we need 9 spots to place the letters. Notice some of the letters repeat. We need to use R 3 times, A 2 times, E 2 times and N and G once. 9 C 6 C 4 C 2 C 1 1 C 1 3 2 2 First we choose positions for the R's. There are 9 positions and we'll choose 3, order doesn't matter 84 15 6 2 1 = 15 120 possible "words" That leaves 6 positions for 2 A's. That leaves 4 positions for 2 E's. That leaves 2 positions That leaves for the N. 1 position for the G.
This can be generalized into the following: Permutations Involving n Objects That Are Not Distinct
Pe A Challenging Example. Have a go. Or rm de uta r M tio How many even numbers greater than 4000 can be at n: formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each ter s digit must feature no more than once in a number? We could have even numbers with 4, 5 or 6 digits This Gives 4 possibilities to work with: PART A: 4, 5 or 6 EVEN digits beginning with a 4 OR 6 PART B: 4, 5 or 6 EVEN digits beginning with a 5 PART C: 5 or 6 EVEN digits beginning with a 2 PART D: 5 or 6 EVEN digits beginning with a 1 or 3
Pe A Challenging Example. Have a go. Or rm de uta r M tio How many even numbers greater than 4000 can be at n: formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each ter s digit must feature no more than once in a number? PART A: 4, 5 or 6 EVEN digits beginning with a 4 OR 6 2 4 + 2 3 4 2 3 + 2 2 4 1 This gives a total of 240 3 2 2 2
Pe A Challenging Example. Have a go. Or rm de uta r M tio How many even numbers greater than 4000 can be at n: formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each ter s digit must feature no more than once in a number? PART B: 4, 5 or 6 EVEN digits beginning with a 5 1 4 + 1 3 4 3 3 + 2 1 4 1 This gives a total of 180 3 3 2 3
Pe A Challenging Example. Have a go. Or rm de uta r M tio How many even numbers greater than 4000 can be at n: formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each ter s digit must feature no more than once in a number? PART C: 5 or 6 EVEN digits beginning with a 2 1 4 1 3 2 4 3 2 + 2 1 This gives a total of 96 2
Pe A Challenging Example. Have a go. Or rm de uta r M tio How many even numbers greater than 4000 can be at n: formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each ter s digit must feature no more than once in a number? PART D: 5 or 6 EVEN digits beginning with a 1 or 3 2 4 2 3 2 4 3 3 + 2 1 This gives a total of 288 3
Pe A Challenging Example. Have a go. Or rm de uta r M tio How many even numbers greater than 4000 can be at n: formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each ter s digit must feature no more than once in a number? We could have even numbers with 4, 5 or 6 digits This Gives 4 possibilities to work with: PART A: 4, 5 or 6 EVEN digits beginning with a 4 OR 6 = 240 PART B: 4, 5 or 6 EVEN digits beginning with a 5 = 180 PART C: 5 or 6 EVEN digits beginning with a 2 = 96 PART D: 5 or 6 EVEN digits beginning with a 1 or 3 =288 Number of possible even numbers greater than 4000 = 804
Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this Power. Point. www. slcc. edu Shawna has kindly given permission for this resource to be downloaded from www. mathxtc. com and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www. ststephens. wa. edu. au
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