1. Permutations and Combinations
Key Vocabulary:
Permutation
Combination
Fundamental Counting Principle
Factorial

2. Objectives: apply fundamental counting principle compute permutations
compute combinations
distinguish permutations vs combinations

3. 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

4.  Example: 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

5. 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, 

6. 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

7. Permutations Notice, ORDER MATTERS!
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.

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

9. 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

10. 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

11. 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.

12. 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

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

14. Permutations Practice:
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?
Answer Now

15. Permutations Practice:
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?

16. Permutations Practice:
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?
Answer Now

17. Permutations Practice:
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?

18. Combinations ORDER DOES NOT MATTER!
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.

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

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

21. 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.

22. 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?

23. Combinations
Practice:
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?
Answer Now

24. 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:

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

26. 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:

27. Combinations
Practice:
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?
Answer Now

28. Combinations Practice:
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?
Center:
Forwards:
Guards:
Thus, the number of ways to select the starting line up is 2*10*6 = 120.

29. Review/Summary Permutations vs. Combinations When is it a permutation?
When is it a combination?

30. A permutation is an arrangement in which order matters.
A B C differs from B C A

31. How Many Permutations? Consider four objects {A,B,C,D}
There are 4 choices for the first slot.
There are 3 choices for the second slot.
There are 2 choices for the third slot.
There is 1 choice for the last slot.

32. 4 x 3 x 2 x 1 = 24 Permutations ABCD ABDC ACBD ACDB ADBC ADCB
BACD BADC BCAD
BCDA BDAC BDCA
CABD CADB CBAD
CBDA CDAB CDBA
DABC DACB DBAC
DBCA DCAB DCBA

33. Generalization There are 4! ways to arrange 4 items.
There are n! ways to
arrange n items.

34. Permutation Example Selecting 3 items out of a set of 5
We have 5 choices for the first item.
We have 4 choices for the second item.
We have 3 choices for the third item.
5 x 4 x 3 = 60 Permutations

35. Permutation Formula

36. Combinations are arrangements in which order does NOT matter.
A, B, C is the same as B, C, A

37. Evaluating
In how many ways may 3 items be selected from a set of 5 without regard to order?

38. We already know that there 60 permutations of these items.
For each set of three, there are
3! or 6 arrangements.
A B C A C B B A C
B C A C A B C B A
All of these are really the same.

39. Consider the set {A,B,C,D,E} These are the combinations.
Our actual answer is 10.
Consider the set {A,B,C,D,E}
These are the combinations.
A,B,C A,B,D A,B,E A,C,D A,C,E A,D,E B,C,D B,C,E B,D,E C,D,E

40. Combination Formula

41. 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

42. 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

43. 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

44. 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

45. 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

46. 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