Lecture 1.ppthffre6dyfut8tigigigigigighh

Permutation narayana

1. PERMUTATIONS
PERMUTATIONS

2. PERMUTATIONS
INTRODUCTION

3. PERMUTATIONS
INTRODUCT
ION
 Fundamental Principle of Counting plays a very
important role in the theory of Permutations.
Before going to
discuss
‘Permutations’ let us
learn the
fundamental
Principle of Counting.

4. PERMUTATIONS
FUNDAMENTAL (COUNTING)
PRINCIPLE
PRODUCT
RULE
We can extend this principle to a finite number of
operations.
If an operation can be performed in ‘m’ ways and second
operation can be performed in ‘n’ ways corresponding to each
performance of the first operation, then the two operations in succession
can be performed in ‘mn’ ways.

5. PERMUTATIONS
If in k operations, the first operation can be performed in
n
1
ways, the second operation can be performed in
n
2
ways,

third operation can be performed in n
3 ways and so on and the
k
th
operation can be performed in n
k ways. Then the k operations in
succession can be performed in n
1n
2n
3…….n
k ways.
Let us see one
example based on
fundamental
principle of
counting

6. PERMUTATIONS
1. A man has 2 different coloured pants and 2 different
coloured shirts. The number of ways that he can select
a pair (i.e. one pant and one shirt) is 2  2 = 4 as
explained below:
Example
s
Let the pants
be:
and shirts
be:
Possibilitie
s :
P
1
P
2
,
S
1
S
2
,
(P
1
S
1
)(P
1S
2 )(P
2
S
1
)(P
2
S
2
), , ,

7. PERMUTATIONS
Solutio
n
The six selections
are
M
1
P
1

C
1,

2. In a shelf there are 3 different Mathematics books, 2
different Physics books and 1 Chemistry book. A student
can select a Mathematics book, a Physics book and
a Chemistry book in 321=6 ways as explained below:
Let the Mathematics books be M
1
,
M
2, M
3 ; Physics books be P
1
,
P
2
;
and Chemistry book be
C
1
.
M
1
P
2

C
1,

M
2P
1
C
1, M
2
P
2
C
1
,
M
3P
1 C
1,

M
3
P
2
C
1

8. PERMUTATIONS
ADDITION
PRINCIPLE
If an operation can be performed in m ways
and second operation can be performed in n ways. , then either of these
two operations can be performed independently in(m + n )
ways.
We can extend this principle to a finite number of
operations also.
This principle is
used for infinite
number of
operations.

9. PERMUTATIONS
If an operation can be performed in a
1
ways,
second operation can be performed in a
2
ways, third operation can be performed in
a
3 waysand so on and the n
th
operation can be performed
in a
n
ways.
Then
anyone of these n operations can be performed in (a
1 + a
2
+ …. + a
n) ways.
Let us see one
example based on
addition principle

10. PERMUTATIONS
M
1, M
2, M
3, C
1,
C
2.
Example: If a shelf contains 3 different Mathematics books
and 2 different Chemistry books., the number of ways to
select one book of any subject either Mathematics or
Chemistry is 3+2=5 as explained below:
The 5 ways
are
Let the Mathematics books be M
1,
M
2
, M
3
;
Solutio
n
Chemistry books be
C
1
, C
2
.
Here, we select
one book of any
subject

11. PERMUTATIONS
PERMUTATI
ON
Each of the
different
There are two types of
Permutations
1. Linear
Permutation
2. Circular
Permutation
Firstly, we
discuss linear
permutation
arrangements
or all of finite set of objects (elements)
is called a
which can be made by taking some
permutation.

12. PERMUTATIONS
A Permutation is said to be a linear permutation if the
objects are arranged in a line or in a row.
A linear permutation is simply written as a permutation.
LINEAR
PERMUTATION

13. PERMUTATIONS
Examples
1. The linear permutations formed by taking two things at
a time from the set {A, B, C} areAB, BA
,
AC, C
A
,

BC
,
CB
.

2. The linear permutations formed by taking two digits at
a time from the set {1, 2, 3, 4} are
Total number of linear
permutations = 6
Can you guess
the answer?
12, 21, 13, 31, 14, 41, 23, 32, 24, 42, 34, 43
Total number of linear
permutations = 12

14. PERMUTATIONS
Not
e
The permutation involves two
steps .
1)
Selection
2)
Arrangement
In the first stage we select the
objects
and in the second
stage we arrange the selected objects in different order.
In permutation, the order of the objects (elements) in
which they are arranged is important.

15. PERMUTATIONS
The number of permutations of n distinct things taken r at
a time is denoted by
n
r
or P
Not
e
n
p
r
or P(n, r)
Here 0 ≤ r ≤ n, n is a positive integer and r is a
positive integer
We write
n
P
0
= 1 by
convention
Let us see one
important formula to
find number of
permutations

16. PERMUTATIONS
PERMUTATIONS OF
DISSIMILAR THINGS

17. PERMUTATIONS
FACTORIAL OF A NON-NEGATIVE
INTEGER
i) 0! =
1

ii) If n ≥ 1
then
If n is a non negative
integer
Note
n!
=
and is defined as
follows :
then factorial n is
denoted by
n! nor
n! = n(n-
1)!
n(n-1)!
=
n(n-1)(n-2)!
=……
=n(n-1)(n-2)(n-3)……
3.2.1.
(n-1)! can be
expanded as
(n-1)(n-2)!

18. PERMUTATIONS
Exampl
e
ii) 6! =
i) 4!
=
4.3.2.1
=
24
6.5.4.3.2.1 =720

19. PERMUTATIONS
PERMUTATIONS OF DISSIMILAR
THINGS
Clearly
n
p
r is equal to the number of ways of filling
r blank places
Proo
f
Theorem: If 1 ≤ r ≤ n and r, n are positive
integers, then
n
p
r
= n(n-1)(n-2)….(n-r+1).
which are arranged in a row by n
dissimilar things.

20. PERMUTATIONS
11

2 534 r. . . . . . . .
Consider r blank places in a row as
shown below
The first blank place can be filled by any one of the n
things and hence it can be done in n ways.
To fill the second blank place, there are (n – 1)
things and hence the second blank place can be filled in (n –
1) ways.

21. PERMUTATIONS
Similarly, the third blank place can be filled in (n – 2)
ways and so on.
By counting principle, the r blank
places can be filled in n(n-1)(n-2)
…….. (n-r+1) ways.
Therefore,
n
p
r = n (n -1) (n – 2)….….(n
– r + 1).
The r
th
blank place can be filled in (n –( r - 1)) = (n-
r+1) ways.

22. PERMUTATIONS

n
p
r=n(n-1)(n-2)…….….(n-r+1).
Multiply and divide R.H.S. by (n-r) (n-r-1) ……3.2.1
The number of permutations of n dissimilar things
taken all at a time is
n
p
n = n!.
n(n-1)(n-2)…(n-r+1)
(n-r)(n-r-1)…3.2.1
=
=
observe that
the numerator
is n!
(n-r)(n-r-1)….3.2.1
formula for
n
p
r

23. PERMUTATIONS
Example
s
1)
6
p
3
= =
= 6x5x4 = 120
2)
9
p
2
= =
= 9x8
= 72
=
=

24. PERMUTATIONS
Theorem:
n
p
r
=
(n-1)
p
r
+ r

.
(n-1)
p
(r-1)
R.H.S.=
(n-1)
p
r
+ r

.
(n-1)
p
(r-
1)
Proo
f
(n-1)!
(n-1-r)!
=
(n-1)!
(n-r)!
+r
n
p
r
=
(n-1)!
(n-1-r)!
=
(n-1)!
(n-r) (n-r-1)!
+r
(n-1)!
(n-1-r)!
= [
1+
r
n−r]
(n-1-r)!
= [
n−r+r
n−r]
(n-1)!
(n-1)!
(n-r)(n-r-1)!
=
n!
(n-r)(n-r-1)!
=
n!
(n-r)!
=
=
n
p
r = L.H.S
This can be written as
Take
as common
n

25. PERMUTATIONS
Second
Proof
Now
n
p
r denotes the number of permutations of n
dissimilar things
(1)The permutations that contain a
particular thing.
(2) The permutations that do not contain that
particular thing.
taken r at a
time.
These permutations can be divided into
two parts.

26. PERMUTATIONS
and the remaining things can be filled in
(n-1)
P
(r-1) ways.
in the other (r-1) blank places.
By counting principle, the number of permutations = r.
(n-1)
P
(r-1).
The particular thing can be filled in
r ways
and next fill the other (n-1)
things
1. To find the number of permutations that contain a
particular thing,
At first fill that particular thing in any one of the r blank
places arranged in a row
............r
First ly, we
take r blank
places
How many blank places are remaining?
r-1
How many things are remaining?
n-1

27. PERMUTATIONS
2. To find the number of permutations that do not contain
that particular thing,
Fill the remaining (n-1) things in the r blank places
arranged in a row.
It can be done in
(n-1)
P
r

ways.
Therefore
n
p
r=Sum of the number of permutations in
both the parts
=
(n-1)
P
r + r

.
(n-
1)
p
(r-1).
Contains a particular
thing + does not
contain a particular
thing

28. PERMUTATIONS
Note
1. Included is r.
(n-1)
P
(r-
1)
2. Excluded is
(n-1)
P
r
The number of permutations of n dissimilar things taken
r at a time in which one particular thing always

29. PERMUTATIONS
i) The number of 4 letter words that can be formed using
the letters of the word MIXTURE,
Find the number of 4 letter words that can be formed
using the letters of the word MIXTURE, which i) contain
the letter ‘X’ ii) do not contain the letter ‘X’
Exampl
e
Solution
4.
(7-1)
P
(4-
1)
which contains the letter
‘X’ is
r.
(n-1)
P
(r-1)
=
4.
6
P
3

30. PERMUTATIONS
ii) The number of 4 letter words that can be formed using
the letters of the word MIXTURE, which do not contain
the letter ‘X’
(7-1)
P
4
(n-1)
P
r
=
6
P
4

31. PERMUTATIONS
Thank you…