Permutationssafafhadbfghdsbgsndgsdgad.pptx

Objective/s: After accomplishing the module,
you must be able to:
1. illustrate the permutation of
objects. (M10SP-IIIa-1)
2. derive the formula for finding the
number of permutations of n
objects taken r at a time. (M10SP-
IIIa-2)
3. solve problems involving
permutations. (M10SP-IIIb-1)

A permutation is an arrangement
of objects in a specific order.
If it is an ordered arrangement
of distinguishable objects, then
repetition among objects is not
allowed. To permute objects is
to arrange them in a certain
order. Here, the order of
arrangement of objects is
important. Hence, the
arrangement ABCD is not the same
as the arrangement BCDA. These

In general, in the arrangement of n
objects, there are n operations
involved. The first operation involves
choosing an item for the first position;
the second operation, choosing an
item for the second position; and so
on until the nth operation. From the
Fundamental Counting Principle the
formula becomes
n(n- 1) … 3 2 1
● ● ● ● .

The symbol denotes the permutation of n things
taken r at a time. Two other common notations that
appear in some books are P(n, r) and .
The total number of permutations of n objects taken r at
a time is given by expression
nPr = n(n - 1)(n - 2)...(n - r + 1)

• The 6 permutations of the 3 letters F, B, and O taken
2 at a time are: FB, FO, BO, BF, OF, and OB.
• The 6 permutations of these 3 letters taken 3 at a
time are: FBO, FOB, BFO, BOF, OFB, and OBF.
For instance,
Finding number of permutations of n different objects taken
r at a time is the same as finding the number of ways that
can fill r places when there are n different things choose
from.

Factorial Notations
The product 5 • 4 • 3 • 2 • 1 can be written in abbreviated from
5! (Read as “Five factorial”).
5! = 5 • 4 • 3 • 2 • 1 = 120
7! = 7 • 6 • 5 • 4 • 3 • 2 • 1 = 7 • 6 • 5! = 5,040
4! = 4 • 3 • 2 • 1 = 24
6! = 6 • 5 • 4 • 3 • 2 • 1 = 6 • 5 • 4! = 720
If n is a positive integer, then n factorial denoted by n! is
the product of all integers less than or equal to n. in other
words, n! = n(n – 1)(n – 2)...(2)(1). As a special case, you
define 0!= 1.

Illustrative Example A:
Find the following.
a. 9! – 6!
b. 3! + 2!
c.
= (9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1) – (6 • 5 • 4 • 3 • 2 • 1)
= 362, 880 – 720
= 362, 160
= 3 • 2 • 1 + 2 • 1
= 6 + 2
= 8
¿
7!
7 !3!
¿
1
3 !
¿
1
3∗2∗1
¿
𝟏
𝟔

Illustrative Example A:
Find the following.
d. ¿
7∗6∗5∗4∗3∗2∗1
7∗6∗5∗4 ∗3∗2∗1−3∗2∗1
¿
5040
5040 −6
¿
5040
5034
¿
𝟖𝟒𝟎
𝟖𝟑𝟗
e.
¿
7∗6∗5∗4 ∗3∗2∗1
4∗3∗2∗1
¿
7 !
4 ! ¿7∗6∗5
¿𝟐𝟏𝟎
f. 7! – 0! = 5, 040 – 1 = 5, 039

Linear Permutation
The formula in computing the number of permutations of n
things taken r at a time can be derived from the fundamental
principle of counting. If the things are to be taken r at a time
where r<n, then you have n things to fill up the first place, n – 1
things to fill up the second place, and so on.
You have (n – r + 1) things to choose from to fill up the rth place. Thus ,
The number of ways of different ordered arrangements of n
things taken r at a time is given by
nPr = n(n - 1)(n - 2)...(n - r + 1)
The number of permutation of n things taken n at a time is
given by

Example 1.
The Racing Club organizes a race in which five cars A, B, C, D,
and E are entered.
a. In how many ways can the race be finished if there are
no ties?
No ties is the same as the number of permutations of 5 things
taken 5 at a time. We denote this number by 5P5 where;
Solution:
5P5 = 5!
= 5 • 4 • 3 • 2 • 1
= 120

Example 1.
The Racing Club organizes a race in which five cars A, B, C, D,
and E are entered.
b. In how many ways can the first two positions come in if
there are no ties?
Solution:
Here, we are required to find the number of permutations of
5 things taken two a time. We denote this number by 5P2,
5P2 = 5 • 4
= 20

The Sick Mind Band have 20 songs. At the upcoming Battle of
the Bonds, they will play two songs. In how many different
orders could they perform two of their songs.
Example 2:
Solution:
We are required to find the number of permutations of 20
things taken two a time. We denote this number 20P2.
❑𝟐𝟎 𝑷𝟐=
𝟐𝟎!
(𝟐𝟎− 𝟐)!
=
𝟐𝟎∙𝟏𝟗∙𝟏𝟖!
𝟏𝟖!
=𝟑𝟖𝟎

Circular
Permutations
Circular permutations are the
permutations of objects when
they are arranged in a circular
pattern. An example of this is
the seating arrangement of
people around a circular table.
The main difference between
circular and linear permutations
is that in a circular permutation
there is no starting or ending
point.

Suppose there are four objects a, b, c, d. in linear permutations, they can
be arranged in 4! = 24 differen1t ways. When the four objects a, b, c, and d
are arranged in a circle, you get the following arrangements:
The arrangements shown to the left
are all considered the same because
there is no starting or ending point.
Thus, out of the 4! = 24 different
ways arranging 4 district objects in a
line, there will be 4 such linear
arrangements that will be considered
the same from a circular
permutation’s view. Therefore, the
number of circular permutations of 4
different objects will be .

In general, if there are n different objects, and they are
arranged in a circular fashion, then the total number of
circular permutations of these n objects is equal to the
number of linear permutations of the n objects divided by n.
this is because out of the n! ways of arranging n district
objects in a line, there will be n such linear arrangements that
will be considered the same from a circular permutation’s
view. Hence,
Circular permutations of n objects =

In how many ways can 8 persons be seated around a circular
table?
Example 2.
This is a circular permutations of 8 things.
Solution:
= (8 – 1)!
= 7!
( 𝒏 − 𝟏 ) !
= 5 040

Example 3.
In how many ways can 5 keys be arranged in a key ring?
Solution:
The number of permutations of 5 keys around the key ring is
(𝒏−𝟏) !
𝟐
¿
(5 −1) !
2
¿
4 !
2
¿
4 ∗3∗2∗1
2
¿
24
2
¿𝟏𝟐

Permutations with Repetitions/Things That are Alike
Sometimes, the elements with which we want to form permutations are
not all different as in the case of the word POP. Thus, to find the number
of distinguishable permutations of the word POP, we must take into
consideration that we cannot distinguish between the 2 P’s in any of the
permutation. We know that the permutations of the 3 letters of the
word POP is 3!, but the 2 P’s can be permuted in 2! ways and do not
produce a different result, the number of distinguishable permutations
P is given by
𝟐! 𝑷=𝟑!𝒐𝒓 𝑷 =
𝟑!
𝟐!
=
𝟑∗𝟐∗𝟏
𝟐∗𝟏
=
𝟔
𝟐
=𝟑
The number of permutations of n things where a things are alike, b
things are alike, and c things are alike, and so forth, is

Example 4.
Find the number of distinguishable permutations of the
letters in each word.
a. basketball
Solution:
There is a total of 10! Permutations in the word basketball but
there are 2 b’s, 2 a’s, and 2 l’s, can be permuted 2!, 2!, and 2!
ways in each permutation without altering the result. Hence,
the number of distinguishable permutations is given by
𝑷=
𝟏𝟎!
𝟐!𝟐!𝟐!
=𝟒𝟓𝟑,𝟔𝟎𝟎

Example 4.
Find the number of distinguishable permutations of the
letters in each word.
b. heterogeneous
Solution:
𝑷=
𝟏𝟑!
𝟒!𝟐!
=𝟏𝟐𝟗 ,𝟕𝟐𝟗 ,𝟔𝟎𝟎
There are 13 letters in the word “heterogeneous” but there
are 4 e’s and 2 o’s. Hence, the number of distinguishable
permutations is given by

Example 5.
In how many different ways can the letters O, G, T, I, D, A, and
P, be arranged so that
a. all 3 vowels are always
together?
Solution:
Because all 3 vowels are always together, you group the 3
vowels (O, I, and A) as one. Then permute the group of vowels
and 4 consonants. This can be arranged in 5! ways. In each of
these arrangements, the 3 vowels can be arranged among
themselves in 3! ways. Hence, the required number of ways the
7 letters can be arranged so that all 3 vowels are always
together
5! 3! = 720 ways
∙

Example 5.
In how many different ways can the letters O, G, T, I, D, A, and P
be arranged so that
b. the vowels O and A are not together
Solution:
The total number of permutations of the 7 letters (3 vowels and 4
consonants) without restriction is 7! = 5 040. To compute the number of
permutations where in the vowels O and A are together, you group O and
A together as one. Then permute this group (group with vowels O and A)
and the remaining 5 letters. This can be done in 6! = 720 ways. In each of
these arrangements, the 2 vowels O and A can be arranged in 2! ways.
Number of permutations = 7! – (6!2!) = 3600 ways

Permutationssafafhadbfghdsbgsndgsdgad.pptx

More Related Content

Similar to Permutationssafafhadbfghdsbgsndgsdgad.pptx

Recently uploaded

Permutationssafafhadbfghdsbgsndgsdgad.pptx