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Combinations and
Permutations
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Counting Integers
 To
count how many integers there are
between two integers, follow these rules:
 If exactly one of the endpoints is included,
subtract.
 If both endpoints are included, subtract
and add 1.
 If neither endpoint is included, subtract
and subtract 1 more.
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Combinations and Permutations
 What’s
 If
the Difference?
the order doesn't matter, it is a Combination
 If the order does matter it is a Permutation
 A permutation is an ordered combination
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Types of Permutations
 Repetition
 These
is Allowed
are the easiest to calculate.
 When
you have n things to choose from ... you
have n choices each time!
 When

 In
choosing r of them, the permutations are:
n × n × ... (r times) or n × n × ... (r times) = nr
other words, there are n possibilities for the
first choice, THEN there are n possibilites for the
second choice, and so on, multiplying each time.
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Formula for problems when
Repetition is Allowed and Order
Matters
 So
the formula is simply:
 nr
 where
n is the number of things to choose
from, and you choose r of them
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Types of Permutations
 No
Repetition
 For
these problems, you have to reduce the
number of available choices each time
 If
you do not want to use all of the numbers
in the series use the following equation
 n!/(n-r)!
 Where
n is the number of things to choose from,
and you choose r of them
 No repetition, order matters
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Factorial Signs
 The
factorial function (!) means to
multiply a series of descending natural
numbers.
 Note: it
is generally agreed that 0! = 1
 This helps to simplify equations
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Calculating Number of Jobs
 If
two jobs need to be completed and there
are m ways to do the first job and n ways to
do the second job, then there are m x n
ways to do one job followed by the other.
This principle can be extended to any
number of jobs.
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Combinations without Repetition
Another way to write this equation is:
 Where
n is the number of things to choose
from, and you choose r of them (nCr)
 No
repetition, order doesn't matter
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Combinations with Repetition
Another way to write this equation is:
 Where
n is the number of things to choose
from, and you choose r of them (nCr)
 Repetition
allowed, order doesn't matter
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Lets Practice
 Determine
whether the following situations would require
calculating a permutation or a combination:
 1)

Selecting three students to attend a conference
permutation combination
2) Selecting a lead and an understudy for a school play.
permutation combination
 3)
Assigning students to their seats on the first day of school.
permutation combination
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Lets Practice
 Determine
whether the following situations would require
calculating a permutation or a combination:
 1)

Selecting three students to attend a conference
Combination
2) Selecting a lead and an understudy for a school play.
Permutation
 3)
Assigning students to their seats on the first day of school.
Permutation
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Practice Problem
 The
local Family Restaurant has a daily
breakfast special in which the customer
may choose one item from each of the
following groups:
Break fast Sandwich
Accompaniments
Juice
Egg and Ham
Breakfast potatoes
orange
Egg and Bacon
Apple Slices
Cranberry
Egg and Cheese
Fresh Fruit Cup
Tomato
Pastry
Apple
Grape
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Practice Problem
Break fast Sandwich
Accompaniments
Juice
Egg and Ham
Breakfast potatoes
orange
Egg and Bacon
Apple Slices
Cranberry
Egg and Cheese
Fresh Fruit Cup
Tomato
Pastry
Apple
Grape
 How
 How
many different breakfast specials are possible?
many different breakfast specials without meat
are possible?
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Practice Problem A&E
 How
many different breakfast specials are
possible?
 Basic
counting principle
 Sandwiches x Accompaniments x Juice
 3 • 4 • 5 = 60 breakfast choices
 How
many different breakfast specials without
meat are possible?
 Meatless
means that under Sandwiches there will be
only one choice
 Sandwiches x Accompaniments x Juice
 1 • 4 • 5 = 20 meatless breakfast choices
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Practice Problem 2
 In
how many ways can a group of 5 men
and 2 women be made out of a total of 7
men and 3 women?
 A. 63
B. 90
 D. 45
E. 135
C. 126
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Practice Problem 2 A&E
 Answer:
Option A
 Explanation:
 Required
number of ways
 = (7C5
x 3C2)
 = (7C2
x 3C1)
=
[( 7 x 6) / (2 x 1)] x 3
=
63
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Practice Problem 3
 In
a group of 6 boys and 4 girls, four
children are to be selected. In how many
different ways can they be selected such
that at least one boy should be there?
 A. 159
B. 194
 D. 209
E. None of these
C. 205
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 Answer:
Practice Problem 3 A&E
Option D
 Explanation:
 We
may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3
boys and 1 girl) or (4 boys).
 Required
number of ways
 = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
 = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
 = (24 + 90 + 80 + 15)
 = 209.
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Any Questions?
 It’s
ok if you are still confused
 These
grasp
questions are generally difficult to