PERMUTATIONS

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COUNTING TECHNIQUES
PERMUTATIONS
AND
COMBINATIONS
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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
chicken
soup
salad
hamburger
prawns
DESSERTS
How many different combinations of meals could
you make?
We'll build a tree diagram to show all of the
choices.
?
icecream
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We ended up with 12 possibilities
Notice the number of choices at each branch
2choices
3choices
2choices
soup, chicken, ice cream
soup, chicken, ?
2 ? 3 ? 2 12
soup, prawns, ice cream
soup, prawns, ?
soup, hamburger, ice cream
soup, hamburger, ?
salad, chicken, ice cream
salad, chicken, ?
salad, prawns, ice cream
salad, prawns, ?
Now to get all possible choices we follow each
path.
salad, hamburger, ice cream
salad, hamburger, ?
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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
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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
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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
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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.
Only one possibility for fourth place.
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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
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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.
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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 doesnt matter. You need
2 committee members but it doesn't matter who is
chosen first. How many combinations are there?
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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 ?
R
R
R
A
A
E
E
N
G
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 15 120 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.
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This can be generalized into the following
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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
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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
4
2
2
2
4

2
2
3
3
2
2
4
3
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
4
3
3
1
4

1
2
3
3
1
3
4
3
1
2

This gives a total of 180
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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
2
1
4
2
3

1
2
4
3
1
2
This gives a total of 96
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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
3
2
4
2
3

2
3
4
3
1
2
This gives a total of 288
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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
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Acknowledgement I wish to thank Shawna Haider
from Salt Lake Community College, Utah USA for
her hard work in creating this PowerPoint. www.sl
cc.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
Stephens School Carramar www.ststephens.wa.edu.
au