Math 30: Pre-Calculus - PowerPoint PPT Presentation

1 / 52
About This Presentation
Title:

Math 30: Pre-Calculus

Description:

F. How Many Ways can You Organize? Math 30: Pre-Calculus PC30.12 Demonstrate understanding of permutations, including the fundamental counting principle. – PowerPoint PPT presentation

Number of Views:299
Avg rating:3.0/5.0
Slides: 53
Provided by: DwightWr
Category:

less

Transcript and Presenter's Notes

Title: Math 30: Pre-Calculus


1
F. How Many Ways can You Organize?
  • Math 30 Pre-Calculus
  • PC30.12
  • Demonstrate understanding of permutations,
    including the fundamental counting principle.
    PC30.13
  • PC30.13 Demonstrate understanding of combinations
    of elements, including the application to the
    binomial theorem.

2
Key Terms
3
(No Transcript)
4
1. Permutations
  • PC30.12
  • Demonstrate understanding of permutations,
    including the fundamental counting principle.

5
1. Permutations
6
  • Counting Methods are used to determine the number
    of members from a set as well as the outcome of
    an event.
  • There are methods such as tables, lists or tree
    diagrams that allow you to visually see all the
    outcomes.
  • Another methods for determining the number of
    outcomes is called the Fundamental Counting
    Principle.

7
  • The fundamental counting principle states that if
    one task can be performed in a ways and second
    task can be performed in b ways, then the two
    tasks can be performed in (a)(b) ways.

8
Example 1
9
  • In the last example when we arranged the 5
    students in the middle we end up with
    (5)(4)(3)(2)(1) ways.
  • This can be abbreviated as 5! and is read as 5
    factorial.
  • Therefore, 5! (5)(4)(3)(2)(1)

10
(No Transcript)
11
  • Seven different objects can arranged 7! Ways
  • If there are 7 members of a student council how
    many ways can they select the chair, secretary
    and treasurer?

12
(No Transcript)
13
Example 2
14
  • With permutations we said order of different
    objects is important. Well what is some of the
    objects in the set are identical.
  • Consider the word WEED and all the possible 4
    letter arrangements.

15
  • If all the letters were different the number of
    outcomes would be 4!24
  • There are however 2 identical letters. If they
    were different we would arrange them 2!2 ways.
  • So the number of arrangements of the word WEED is

16
  • For permutations with repeating objects, a set of
    n objects with a of one kind that are
    identical, b of a second kind that are
    identical, and c of a third kind that are
    identical, and so one, can be arranged in

17
Example 3
18
Example 4
19
  • To solve some problems you must count the
    different arrangements in all the cases that
    together cover all the possibilities.
  • Calculate the number of arrangements for each
    case and then add that values for all cases to
    obtain the total number of arrangements.
  • Whenever you encounter a situation with
    constraints or restriction, always address the
    choices for the restricted positions first.

20
  • For example, you may need to determine the number
    of arrangements of 4 girls and 3 boys in a row of
    7 seats if the end of the rows must be either
    both male or both female.

21
Example 5
22
Key Ideas p. 526
23
Practice
  • Ex. 11.1 (p.524) 1,2-8 odds in each, 9-18 evens,
    22
  • 5-8 odds in each, 9, 8-26 evens

24
2. Combinations
  • PC30.13
  • PC30.13 Demonstrate understanding of combinations
    of elements, including the application to the
    binomial theorem.

25
2. Combinations
26
  • A combination is a selection of a group of
    objects taken from a larger group.
  • The kinds of objects selected is important but
    NOT the order in which they are selected.
  • There are a few ways to find the possible number
    of combinations

27
  • On is to use reasoning. Use the fundamental
    counting principle and divide by the number of
    ways that the objects can be arranged among
    themselves.
  • For example, calculate the number of combinations
    of 3 digits made from 1-5 without repetition.

28
  • There are 60 ways to arrange 3 items form 5
  • However, 3 digits can be arranged 3! Ways among
    themselves and in a combination there are
    considered the same selection.
  • So

29
(No Transcript)
30
  • So number of ways of choosing 3 digits from five
    digits is

31
  • The number of combinations of n items taken r
    at a time is equivilent to teh number of
    combinations of n items taken n-r at a time that
    is nCrnCn-r
  • Proof

32
  • To solve some problems, count the different
    combinations in cases that together cover all the
    possibilities.
  • Calculate the number of combinations for each
    case and then add the values for all cases to
    obtain the total number of combinations.

33
Example 1
34
Example 2
35
Example 3
36
  • When answering questions it is important to know
    if you are dealing with a permutation or a
    combination.
  • Remember in Permutations the order of the objects
    is important.
  • IN the combinations the type of objects is
    important but NOT the order in which they are
    selected.

37
Key Ideas p.533
38
Practice
  • Ex. 11.2 (p.534) 1-6 odds in each, 7-13, 14-20
    evens
  • 4-6 odds in each, 7-13, 14-24 evens

39
3. The Binomial Theorem
  • PC30.13
  • PC30.13 Demonstrate understanding of combinations
    of elements, including the application to the
    binomial theorem.

40
3. The Binomial Theorem
41
  • If you expand a power of a binomial expression,
    such as (xy)4 you get a series of terms
  • There are many patterns in the expression of
    (xy)4

42
(No Transcript)
43
(No Transcript)
44
(No Transcript)
45
Example 1
46
(No Transcript)
47
  • For coefficients you can use Pascals triangle
    instead of Combinations.

48
  • Important Properties of the binomial expansion
    (xy)n include
  • Write binomial expansions in descending order of
    exponent of the first term in the binomial
  • The number of objects, k, selected in the
    combination nCk can be taken to match the number
    of factors of the second variable. That is, it
    is the same as the exponent on the second
    variable.
  • The sum of the exponents in any term of the
    expansion is n.

49
  • The General Term tk1 has the form
  • nCk (x)n-k(y)k

50
Example 2
51
Key Ideas p.541
52
Practice
  • Ex. 11.3 (p.542) 1-7 odds in each, 8-9, 11-19
    odds
  • 2-3, 5-7 odds in each, 9-25 odds
Write a Comment
User Comments (0)
About PowerShow.com