PREFACE
Gog create the world
and everything inside so that we, human , can invited many new creations in
older to help us survive. To survive, human need to have the ability to think
and to learn. Therefore, by learning mathematics, we get much knowledge that
necessary to develop our technology.
Mathematics is a
subject that challenges you to be creative in thinking. Because of that, this
book will nurture and guide you in a right direction to understand the concepts
of mathematics. We us simple and direct choices of language in order to make it
easier for you understand mathematics concept and its application. In the
beginning of every chapter, we will give you an example of what you will learn,
in the form of a simple test question meant as stimulus to interest you in studying further. The examples of test
question and exercise vary ; it meant to check your understanding, knowledge,
application, and analysis. The test questions explanations are done step by
step, with some alternative ways to solve it, to make sure that you understand
the theory application in the form of test question by yourselves.
By studying this
book, we hope that you can learn by yourselves, and apply math in your daily
life. Therefore, we hope that this book can be your friend and your source of
fun in learning mathematics concept.
Permutation of a group of object is different
arrangement from the objects by concerning the order.
PERMUTATION
There is a letter set that is {a, b, c}. Those letters can be arranged in
different order.
Example:
a.
Arrangement of 3 letters, i.e. abc, acb, bac,
bca, cab, and cba.
b.
Arrangement of 2 letters, i.e. ab, ba, ac, ca,
bc, and cb.
The
letters arrangement, both 3 or 2 letters, is called permutation of a, b, and c.
Generally, it
is:
Permutation for
example:
a.
Is called permutation of three of three object
and is symbolized 3 P3 while
permutation for example
b.
Is called permutation of two of three object and
is symbolized 2P2.
The total of permutation of r object which is arranged from n object is
annotated by nPr or
Prn which is
formulated as follows.
nPr =
n(n-1) n(n-2) n(n-3) . . . (n-r+1)
or
nPr =
In the first place or the
permutation can be taken from any object of n
objects. Therefore, there is n way.
The second place is taken by any objects except one that has been used in the
first place. So there are (n-1) ways.
For the third place are (n-2) ways, the forth place there are (n-3)ways, and so on. Thus for
the rth place, the
are (n (n-1)) = (n –r+1) ways. Based
on the multiplication rule, there will be n(n-1)(n-2)
. . . (n-(r-1)) or nPr
=
=
For
n =r we will got
nPr =
=
=
= n!
Therefore
for example (a),
3P3 =
=
=
3!= 3 x 2 x 1 = 6
For
example (b),
3P2
=
=
=
=
6
Example:
Calculate:
a. 5P5 b. 5P4
Solution :
a. 5P5
= 5! = 5 x 4 x 3 x 2 x 1 =120
b. 5P4
=
=
=
5! = 5 x 4 x 3 x 2 x 1 = 120
A.
Permutation With Some Indetical Objects
For letters ABA how
many ways can be arranged? The arranged is
ABA, AAB, and BAA.
Therefore,
there are 3 ways.
If we us permutation formula,
that is 3P3 then we get 6 ways 3P3
= 3! = 3 x 2 x 1 = 6 (by differentiating
letter A). Suppose, letter A is differentiated into A1 and A2
then we get arrangement: A1BA2, A1A2B,
BA1A2, A2BA1, A2A1B,
and BA2A1.
In reality, letter A is not
differentiated. It means there are 2 same arrangements, i.e.
A1BA2 = A2BA1
A1A2B = A2A1B
BA1A2 = BA2A1
So the are only 3 arrangement.
3P2
=
=
=
3
For the explanation above, we
can conclude as follows.
a.
The total of permutation from n object with x same object ( x-n ) is nPr =
b.
The total of permutation from n object
where the are several same objects. Suppose m1, the same object,
there is m2 the same object, there is m3 the same object,
and so on.
nPm1
, m2 , m3 , . . . =
Example:
How many permutation are made
from word “ANANTA”?
Solution
:
From 6 letters of word “ANANTA”,
the are 3 A letter, 2 N letter, and 1 T letter. Therefore the number of
permutation is:
5P3,2 =
=
=
= 60
B.
Cyclic Permutation
Look
at the figure A, B, and C is arrangement in circle. If the arrangement is
suppose to have the same direction as the clock goes, then the arrangement is
ABC, CAB, and BCA is the same. If the arrangement is supposed not to have same
direction as the clock goes, then the arrangement is CBA, BAC, and ACB is the
same. Then, the number o permutation of 3 object are =
=
=
2! = 2.
Therefore,
2 different arrangement are ABC and ACB.
Permutasion of 3 object in circle.
Circle permutasion with 4 object.
The
four object of figure above show the same permutation. So the number of
permutation are
=
=
3!
It means, the number different
arrangement is 3! = 3 x 2 x 1 = 6
In generally we can be conclude that. The
number of cyclic permutation of n
object = (n – 1).
Daftar Pustaka
Sunardi, hari Subagya.2011.Students Guide to Understanding Mathematics
SMA/MA. Jakarta:PT Bumi Aksara.
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