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Mathematics
Session
Permutation & Combination
Session Objective
1. Factorial
2. Fundamental principles of counting
3. Permutations as arrangement
4.
nP
r
formula
5. Permutations under conditions
6. Permutation of n objects taken r at a time
Permutation – Its arrangement
Two element –
a
b
Arrangements  (a, b), (b, a) = 2
Son
Father
Father is riding
Father Son
Father is teaching
Permutation – Its arrangement
Three elements a, b, c
Arrangements : First Second Third
Place Place
Place
a
b
c
a
c
b
b
a
c
b
c
a
c
a
b
c
b
a
=6
Permutation – Its arrangement
Ist
3 ways
2nd
2 ways
3rd
1 ways
Total ways = 3 + 2 + 1 ?
or
3x2x1?
Permutation – Its arrangement
Ist
2 ways
2nd
1 ways
Total ways = 2 + 1 = 3
or
2x1=2
x
Permutation – Its arrangement
Ist
3 ways
2nd
2 ways
3rd
1 ways
Total ways = 3 + 2 + 1 ?
or
3x2x1?
x
Number of Modes A  B?
Cycle
Scooter A
B
1 Km
Car
Add/Multiply
Number of modes to reach B = 1 +1 + 1 + 1 = 4
Car Walking
Bus
Scooter
Number of ways A  B?
I
A
II
Independent Process
B
No. of ways = 1 + 1 = 2
Mode & Way
Number of style A  B?
Scooter
Cycle
I
A
II
B
Car
Ways – 2
Modes – 4
To reach B, one dependent on both ways and mode.
Number of style = 4 x 2 = 8
Independent Process  +
Dependent Process  X
Mode & Way
There are two ways and 4 modes for
A  B. How many way one can reach
B from A?
One can reach Lucknow from New Delhi
only through Kanpur (No direct root)
New
Delhi
I
II
III
IV
A
Kanpur
Process  Dependent
No. of ways = 4 x 2 = 8
B
Lucknow
IA
IB
II A
II B
III A
III B
IV A
IV B
Mode & Way
New
Delhi
I
II
III
IV
IV
Kanpur
V
A
Lucknow
B
Questions
Illustrative Problem
From the digits 1, 2, 3, 4, 5 how many two
digit even and odd numbers can be formed.
Repetition of digits is allowed.
Solution:
Total nos = 5
Even number
Even numbers=5 x 2=10
5 ways 2 ways (2/4)
Solution contd..
Even Numbers=10
Odd number
5 ways 3 ways (1/3/5)
Odd numbers = 5 x 3 = 15
Total numbers
Total numbers = 5 x 5 = 25
5 ways
5 ways
Illustrative Problem
From the digits 1, 2, 3, 4, 5 how many two
digit numbers can be formed. When
repetition is not allowed.
Solution:
5 ways
Total = 5 x 4 = 20
4 ways
Illustrative Problem
There are three questions. Every question can
be answered in two ways, (True or False). In
how many way one can answer these three
questions?
Solution:
Question
Ist
2nd
3rd
T
T
F
T
F
2 ways
(T/F)
2 ways
No. of ways = 2 x 2 x 2 = 8
2 ways
F
T
F
T
F
T
F
T
F
Illustrative Problem
In a class there are 10 boys and 8 girls.
For a quiz competition, a teacher how
many way can select
(i) One student
(ii) One boy and one girl student
Solution:
Girl
Boy
8ways
10 ways
(i) Independent of whether boy/girl = 10 + 8 = 18
ways
(ii)Dependent process = 10 x 8 = 80 ways
Illustrative Problem
In a class there are 10 boys and 8 girls.
For a quiz competition, a teacher how
many way can select
(iii) two boys and one girl
(iv) two students
Solution : (iii)
Girl
Boy1
10
8
(iv)student=10+8=18
student1
9
Boy2
10 x 9 x 8 = 720
student2
=18 x 17
18
17
Illustrative Problem
In a class there are 10 boys and 8 girls.
For a quiz competition, a teacher how
many way can select
(v) at least one girl while selecting 3
students
Solution: case1: 1 girl
G
B
B
boys-10
girl-8
10
8
9 10 x 9 x 8 = 720
G
G
B
case2: 2 girl
case3: 3 girl
8
7
10
8 x7 x 6 = 186
Ans: 720+560+186=1666
8 x 7 x 10 = 560
Principle of Counting
Multiplication Principle : If a job can be
done in ‘m’ different ways, following which
another can be done in ‘n’ different ways and
so on. Then total of ways doing the jobs
= m x n x …… ways.
Addition Principle : If a job can be
done in ‘m’ different ways or ‘n’ different ways
then number of ways of doing the job is (m + n).
Multiplication – Dependent Process
Addition
– Independent Process
Questions
Illustrative Problem
Eight children are to be seated on a bench.
How many arrangements are possible if the
youngest and eldest child sits at left and right
corner respectively.
Solution:
We have 6 children to be seated
Youngest
Eldest
6
5
4
3
2
1
No. ways = 6 x 5 x 4 x 3 x 2 x 1 = 720 x 2 = 1440 ways
Illustrative Problem
A class consists of 6 girls and 8 boys. In how
many ways can a president, vice president,
treasurer and secretary be chosen so that
the treasurer must be a girl and secretary
must be a boy. (Given that a student can’t
hold more than one position)
Solution :
Girls – 6
Boys - 8
Treasurer (Girl)
Girls – 5
Boys - 8
Secretary (Boy)
6 ways
8 ways
Solution contd..
Treasurer(Girl)-6 ways
Secretary(Boy)-8 ways
Girls – 5
Boys – 7;
Total = 12
President
Vice President
12 ways
11 ways
Total = 6 x 8 x 12 x 11
Factorial
•
•
Defined only for non-negative integers
Denoted as n ! or n .
N ! = n . (n - 1) . (n – 2) …… 3 . 2 . 1.
Special case : 0 ! = 1
Example :
3!
=3.2.1=6
5!
= 5 . 4 . 4 . 3. 2 . 1 = 120
(4.5) ! - Not defined
(-2) ! - Not defined
Questions
Illustrative Problem
1
1
x
If

=
then find x.
9! 10! 11!
Solution :
11! 11!
x=

9! 10!
11  10  9! 11  10!
=

9!
10!
= 110  11 = 121
Illustrative Problem
Find x if 8! x! = (x + 2)! 6!
Solution :
(x  2)! 8!
=
x!
6!
(x  2)(x  1)x! 8.7.6!

=
x!
6!
 (x  2)(x  1) = 8.7 = 56
Short
cut
 x2  3x  2 = 56
 x2  3x  54 = 0
 x = 9, 6
As x  0  x = 6
Permutation
Arrangement of a number of object(s) taken
some or all at a fine.
Example :
Arrangement of 3 elements out of
5 distinct elements = 5.4.3 =
=
5.4.3.2.1 5!
=
2.1
2!
5!
=5 P3
(5  3)!
nP
r
Arrangement of r elements out of n given
distinct elements.
1st
2nd
rth
…….
?
n
(n - 1)
(n – r + 1)
No. of arrangement =
n.(n  1)(n  2)....(n  r  1)
nP
r
No. of arrangement =
?!
n.(n  1)(n  2)....(n  r  1) =
?!
n.(n  1)(n  2)....(n  r  1)(n  r)....2.1
=
(n  r)(n  r  1)...2.1
n!
=
=n Pr
(n  r)!
(r  n)
For r = n
Arrangements of n distinct element nPn = n!
taken all at a time.
Questions
Illustrative Problem
In how many way 4 people (A, B, C, D)
can be seated
(a) in a row
(b) such that Mr. A and Mr. B always sit
together
Solution :
(a) 4P4 = 4!
(b) (A, B), C, D
treat as one
Solution contd..
(b) (A, B),
C, D
Arrangement among 3 = 3P3
3P
3
(A, B), C, D
(A, B, C, D)
(B, A, C, D)
2P
2
(A, B), D, C
(A, B, D, C)
(B, A, D, C)
2P
2
C, (A, B), D
C, D, (A, B)
D, (A, B), C
D, C, (A, B)
3P
3
x 2P2
= 3!2! = 12
Solution contd..
In how many way 4 people (A, B, C, D)
can be seated
(c) A,B never sit together
= Total no. of arrangement – No. (A, B) together
= 4P4 – 3P3.2P2
= 4! - 3! 2!
= 24 – 12 = 12
Illustrative Problem
Seven songs (Duration – 4, 4, 5, 6, 7,
7, 7, 7 mins.) are to be rendered in a
programme
(a) How many way it can be done
(b) such that it occurs in ascending order
(duration wise)
Solution :
(a) 7P7 = 7!
(b) Order – 4, 4,
2P
2
5, 6,
7, 7, 7 mins
3P
3
 No. of way = 2P2 x 3P3 = 2! 3! = 24
Illustrative Problem
How many four digits number can be
formed by the digits. 3, 4, 5, 6, 7, 8
such that
(a) 3 must come
(b) 3 never comes
(c) 3 will be first digit
(d) 3 must be there but not first digit
Solution
(a) 3, 4, 5, 6, 7, 8
Digits
available
Position
5
5
5
5
3
3
3
3
Arrangements
3   
 3  
  3 
   3
5
P3
5
P3
5
P3
5
P3
No. of 4 digit numbers with 3 = 4 x 5P3
‘3’ can take any of four position.
In each cases. 5 digits to be arranged in 3 position.
Solution contd..
(b) Digits available – 5 (4, 5, 6, 7, 8
No. of 4 digit numbers without 3 = 5P4
(c) 3 _ _ _
No. of digits available = 5
No. of position available = 3
No. of 4 digit number start with ‘3’ = 5P3
Solution contd..
(d) 4 digit nos. contain ‘3’ but not at first
= 4 digit number with ‘3’ – 4 digit
number with ‘3’ at first
= solution (a) – solution (c)
= 4.5P3 – 5P3 = 3.5P3
Illustrative Problem
In how many way a group photograph of
7 people out of 10 people can be taken.
Such that
(a) three particular person always be there
(b) three particular person never be there
(c) three particular always be together
Solution:
(a) 3 particular 7 places 
With each arrangement
X _
X _ X _ _
Arrangements
7P
3
Solution contd..
Arrangements
Person available – 7
Places available – 3
7P
4
Total no. of arrangements =
Person available = 7
Places available = 7
arrangements
7P
3
x 7P4
7P
7
Solution contd..
(c) X
X
X ____
Treat as one
No. of person = 10 – 7 + 1 = 8
Place available = 5
of which one (3 in 1) always be there.
 No. of arrangement = 5.7P4
3 Particular can be arranged = 3P3 way
 Total arrangement = 5.7P4.3P3
Illustrative Problem
How many way, 3 chemistry, 2 physics,
4 mathematics book can be arranged such
that all books of same subjects are kept
together.
Solution:
Chemistry
1
2
3
Mathematics
Physics
1
2
1
2
3
Intra subject
Arrangements
3P
3
2P
2
4P
4
4
Solution contd..
Inter subject arrangement
Phy
Chem
Maths
 Arrangement = 3P3
Total no. of arrangements = 3P3(3p3 x 2p2 x 4p4)
Class Test
Class Exercise - 1
n!
n!
and
If
are in the
2!  n  2 !
4! n  4 !
ratio 2 : 1, find the value of n.
Solution
Given that
4! n  4 !
n!

=2
2! n  2 !
n!
43
4  3  2!  n  4 !

=2

=2
n  2n  3
2! n  2 n  3 n  4 !
 n2  5n  6 = 6  n2  5n = 0  n n  5 = 0
 n = 0 or n = 5.

Answer is n = 5 (rejecting n = 0).
Class Exercise - 2
How many numbers are there between
100 and 1000 such that each digit is
either 3 or 7?
Solution
By fundamental principle of counting,
the required number = 2 × 2 × 2 = 8
(Each place has two choices.)
Class Exercise - 3
How many three-digit numbers can be
used using 0, 1, 2, 3 and 4, if
(i) repetition is not allowed, and
(ii) repetition is allowed?
Solution
(i) Hundred’s place can be filled in 4 ways.
Ten’s place can be filled in 4 ways.
Unit’s place can be filled in 3 ways.
Required number = 4 × 4 × 3 = 48
(ii) Similarly, the required number = 4 × 5 × 5 = 100
Class Exercise - 4
How many four-digit numbers have at
least one digit repeated?
Solution
Number of four-digit numbers
= 9 × 10 × 10 × 10 = 9000
Number of four-digit numbers with no repetition
= 9 × 9 × 8 × 7 = 4536
 Number of four-digit numbers with at least
one digit repeated = 9000 – 4536 = 4464
Class Exercise - 5
There are 5 periods in a school and 6
subjects. In how many ways can the
time table be drawn for a day so that
no subject is repeated?
Solution
Six subjects can be allocated to five periods in 6 p9 = 6!
ways, without a subject being repeated.
Class Exercise - 6
Number of ways in which 7 different
sweets can be distributed amongst
5 children so that each may receive
at most 7 sweets is
(a) 75
(c) 7p5
(b) 57
(d) 35
Solution
Each sweet can be given to any of the 5
children.
Thus, the required number is
5 × 5 × 5 × 5 × 5 × 5 × 5 = 57
Hence answer is (b)
Class Exercise - 7
In how many ways can 5 students
be seated such that Ram always
occupies a corner seat and Seeta
and Geeta are always together?
Solution
Ram can be seated in 2 ways. Seeta
and Geeta can be together in 3 ways.
(If Ram occupies seat 1, Seeta-Geeta
can be in 2-3, 3-4 or 4-5.)
Seeta and Geeta can be arranged in 2 ways.
Remaining students can be arranged in 2
ways.
Total ways = 2 × 3 × 2 × 2 = 24
Class Exercise - 8
How many words can be made from
the word ‘helicopter’ so that the
vowels come together?
Solution
Treating the vowels as one unit, we have 7 units.
These can be arranged in 7! ways.
The vowels can be arranged in 4! ways.
Total ways = 7! × 4! = 120960
Class Exercise - 9
There are 5 questions. Each question has
two options (one answer is correct). In
how many ways can a student fill up the
answer sheet, when he is asked to
attempt all the questions?
Solution
Every question can be answered in two ways.
 Five questions can be answered in
= 2 ×2 × 2 × 2 × 2 = 25 = 32 ways.
Class Exercise - 10
In how many ways can 6 students (3 boys
and 3 girls) be seated so that
(i) the boys and girls sit alternatively,
(ii) no 2 girls are adjacent?
Solution
(i) Case 1: From left-side, when first student is
a boy, then the boys can occupy Ist, 3rd and
5th places. And the girls can occupy 2nd, 4th
and 6th places.
So the boys can be seated in 3p3 ways and the
girls can be seated in 5p3 ways.
Number of arrangement = 3! . 3!
Solution contd..
Case 2: When the first student is a girl
(from left), then also the number of permutation = 3! × 3!
Therefore, total number of permutation = 2 × (3!)2 = 72
(ii) Let first boys are seated.
They can sit in three places in 3p3 = 3! ways. Since no girls
should be adjacent, the number of seats left for girls are four.
__ B __ B __ B __
Number of permutation for girls = 4p3 = 4!
Therefore, total number of permutation = 3! 4! = 144
Thank you