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COUNTING TECHNIQUES PERMUTATIONS AND COMBINATIONS.

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1 COUNTING TECHNIQUES PERMUTATIONS AND COMBINATIONS

2 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 chicken prawns hamburger DESSERTS How many different combinations of meals could you make? We'll build a tree diagram to show all of the choices. icecream

3 Now to get all possible choices we follow each path. 
We ended up with 12 possibilities Notice the number of choices at each branch 2 choices 3 choices 2 choices ice cream soup, chicken, ice cream chicken 2  3  2 = 12 soup, chicken,  ice cream prawns soup, prawns, ice cream soup hamburger soup, prawns,  ice cream soup, hamburger, ice cream soup, hamburger,  salad ice cream salad, chicken, ice cream chicken salad, chicken,  ice cream prawns salad, prawns, ice cream hamburger salad, prawns,  ice cream Now to get all possible choices we follow each path. salad, hamburger, ice cream salad, hamburger, 

4 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 different ways. p  q  r   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  4  5  3 = 360

5 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 choosing r of them you can repeat an object. this means different Let's look at a 3 combination lock with numbers 0 through 9 There are 10 choices for the first number There are 10 choices for the second number and you can repeat the first number There are 10 choices for the third number and you can repeat By the multiplication principle there are 10  10  10 = 1000 choices

6 Permutations: Distinct Objects with Repetition
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 = choices

7 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: 4  3  2  1 = 24 choices First place would be choosing someone from 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. 4th 3rd 2nd 1st Only one possibility for fourth place.

8 nPr , means the number of ordered arrangements of r objects chosen from n distinct objects and repetition is not allowed. In the last example: If you have 10 people racing and only 1st, 2nd and 3rd place how many possible outcomes are there? 0! = 1

9 A combination is an arrangement of r objects chosen from n objects regardless of order.
nCr , means the number combinations of r objects chosen from n distinct objects and repetition is not allowed. 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.

10 You need 2 people on your committee and you have 5 to choose from
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?

11 The third type of permutation is involving n objects that are not distinct.
How many different combinations of letters in specific order (but not necessarily English words) can be formed using ALL the letters in the word REARRANGE? representative example 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. 9C3  6C2  4C2  2C1  1C1 84  15  6  2  1 = possible "words" First we choose positions for the R's. There are 9 positions and we'll choose 3, order doesn't matter That leaves 6 positions for 2 A's. That leaves 2 positions for the N. That leaves 1 position for the G. That leaves 4 positions for 2 E's.

12 Permutations Involving n Objects That Are Not Distinct
This can be generalized into the following: Permutations Involving n Objects That Are Not Distinct

13 A Challenging Example. Have a go. 
Permutation: Order Matters How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each 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

14 A Challenging Example. Have a go. 
Permutation: Order Matters How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each 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 3 2 + 2 4 3 2 2 + 2 4 3 2 1 2 This gives a total of 240

15 A Challenging Example. Have a go. 
Permutation: Order Matters How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number? PART B: 4, 5 or 6 EVEN digits beginning with a 5 1 4 3 3 + 1 4 3 2 3 + 1 4 3 2 1 3 This gives a total of 180

16 A Challenging Example. Have a go. 
Permutation: Order Matters How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number? PART C: 5 or 6 EVEN digits beginning with a 2 1 4 3 2 2 + 1 4 3 2 1 2 This gives a total of 96

17 A Challenging Example. Have a go. 
Permutation: Order Matters How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each 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 3 2 3 + 2 4 3 2 1 3 This gives a total of 288

18 A Challenging Example. Have a go. 
Permutation: Order Matters How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each 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

19 Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. Shawna has kindly given permission for this resource to be downloaded from and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar

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