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Class 11 Maths Revision Notes Permutations and Combinations

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Mathematics – Iii

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Revision Notes

Class 11 Mathematics

Chapter 7 – Permutations and Combinations

Fundamental principles of counting:
● Fundamental principle of multiplication – If there are three different events
such that one event occurs in m different ways, second event happens in n
different ways and the third event occurs in p different ways, then all three
events simultaneously will happen in m  n p ways.
● Fundamental principle of addition – If there are two jobs such that first wone
can be performed independently in m number of ways and the second work
independently can be done in n number of ways, then wither of the two jobs can

be performed in  m n ways.

Some basic arrangements and selections:
● Permutations – It is the linear arrangements of distinct objects taken some or
all at a time. The number of arrangements possible is called the permutations. If
we have two positive integers r and n such thatl  r n , then the total number
of arrangements or permutations possible for n distinct items taken r at a time
is mathematically given by,
n r!
P
n!
r
n

 

.
● Combinations – If we have to select combinations of items from a given set of
items such that the order or arrangement doesn’t matter, then we use
combinations. Such that to find the number of ways of selecting r objects from
a set of n objects, then mathematically it is given by,
r! n r!
C
n!
r
n

 

Geometric applications of Cr

####### n

:
● If there are n non-parallel and non-concurrent lines, then the number of points
of intersections is given by C 2

####### n

. Similarly, having n points, the number of lines
will be C 2

####### n

such that no three points are collinear.
● If in n number of points, there are m collinear, then number of straight lines is
given by C 2  C 2  1

####### n m

.

● If a polygon has n number of vertices such that no three are collinear, then the

number of diagonals can be given by  

####### 

####### 2

####### C n

####### n n 3

2

n  

.

● If we have n points such that no three are collinear, then the number of triangles

that can be formed is given by C 3

n

.

● If we have n points out of which m are collinear, then the number of triangles

that can be formed is given by C 3  C 3

####### n m

.

● A polygon having n vertices, then the number of triangles that can be formed

such that no side of the triangle is common to that of the side of the polygon is

given by   

C 3 C 1 C 1 C 1

n n n n 4

 .

● Number of parallelograms that can be formed from two sets of parallel lines such

that one set have n parallel lines and other set have m parallel lines is given by

C 2  C 2

n m

.

● Number of squares that can be formed from two sets of equally spaced parallel

lines such that one set have n parallel lines and other set have m parallel lines

is given by    

m r n r ; m n

r 1

m 1

   .

Permutations under certain conditions:

The total number of arrangements or permutations taken r at a time from a set of

n different objects;

● When we always have to include a particular object in every arrangement is

 

C r 1 r!

n 1

.

● When we don’t have to include a particular object in any arrangement is

C r r!

n 1

.

Division of objects into groups:

● Division of items into groups of unequal sizes -

(i) A set of  m n  distinct objects can be divided into two unequal groups

containing m and n objects is given by

m!n!

 m n! 

.

(ii) A set of  m  n p distinct objects can be divided into three unequal groups

containing m,n and p objects is given by 

    

m!n!p!

C. C

m n p!

m m

m n p n p  

.

● Suppose, we have to divide n identical items among r persons, such that each

person receives 0,1,2 or more items until it is  n , then the number of ways it

can be done can be given by 

#######  

####### Cr 1

####### n r 1

. In this distribution, blanks are allowed

means a person may also get no items.

● Suppose, we have to divide n identical items among r persons, such that each

person receives 1,2,3 or more items until it is  0 and  n , then the number of

ways it can be done can be given by 

####### 

####### Cr 1

####### n 1

. In this distribution, blanks are not

allowed means a person has to get at least one item.

● Suppose, we have to divide n identical items among r groups, such that no

group get less than k and more than m items such that  m k , then the number

of ways it can be done is the coefficient of x

n

in the expansion of

  

x x .... x

m m 1 k r

 .

Number of integral solutions of linear equations and inequation:

Let us consider the equation x 1  x 2  x 3  x 4  .....  x r n, where

x ,x ,x ,x ,.....,x 1 2 3 4 r and n are non-negative integers. We can interpret this equation

as n identical objects to be divided into r groups.

● The total number of non-negative integral solutions of the equation

x 1  x 2  x 3  x 4  .....  x r n will be 

#######  

####### Cr 1

####### n r 1

.

● In the set of natural numbers N , the total number of solutions of the equation

will be 

####### 

####### Cr 1

####### n 1

.

● We can solve the equations of the form x 1  x 2  x 3  x 4  .....  x m n by

introducing an artificial or dummy variable xm 1 such that x m 1  0. It will

convert the equation to x 1  x 2  x 3  x 4  .....  x m  x m 1 n and then the

number of solutions can be found out in the same way as given in above points.

Circular permutations:

● If we have n distinct objects, then the number of circular permutations possible

is given by  n 1! .

● If in circular permutations, we also consider the anti-clockwise and clockwise

order as non-distinct, then the number of circular permutations can be given by

2

n 1!

1

  . This can be seen in the arrangement of beads in a necklace or

flowers in a garland, etc.

Selection of one or more objects:

● If we have a group of n distinct items, then the number of ways in which we can
select one or more items from that group is given by 2  1
n
.
● If we have a group of n identical items, then the number of ways in which we
can select r items from that group is always 1.
● If we have a group of n identical items, then the number of ways in which we
can select zero or more items from that group isn  1.

● If we have a group of  p  q r items, such that p items are alike of one kind,

q items are alike of second kind and r items are alike of third kind, then the
number of ways in which we can select some or all out of given items is given

by   p  1  q  1  r  1  1.

● If we have a group of  p  q r items, such that p items are alike of one kind,

q items are alike of second kind and r items are alike of third kind, then the
number of ways in which we can select some or all out of given items is given

by   p  1  q  1  r  1  1.

● If we have a group of  p  q r items, such that p items are alike of one kind,

q items are alike of second kind and r items are alike of third kind, then the
number of ways in which we can select one or more items is given by
 
 p  1 q  1 r  1 2  1
n

   .

Number of divisors and the sum of the divisors of a given natural number:
Let us consider N p 1 .p 2 .p 3 .p 4 .....

####### n 1 n 2 n 3 n 4 nk

, where p ,p ....,p 1 2 k are distinct prime
numbers and n ,n ....,n 1 2 k are positive integers.

● Then, the total number of divisors   n 1  1  n 2  1 ... n  k 1 .

● The above divisors also include 1 and n as divisors, then the number of divisors

other than 1 and n is   n 1  1 n 2  1 ... n  k 1   2.

● And the sum of all divisors is given by
          
      
       
   
p 1 p 1 p 1 p 1
....
p p p p 1
1 2 3 k
1 2 3 k
n 1 1 n 2 1 n 31 n k 1
.
Dearrangements:
● Suppose n objects are arranged in a row, then the number of ways in which we
can arrange the objects so that none of them occupies their original places is
given by D(n) which is equal to
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Class 11 Maths Revision Notes Permutations and Combinations

Course: Mathematics – Iii

39 Documents
Students shared 39 documents in this course

University: MATS University

Was this document helpful?
Class XI Maths 1
Revision Notes
Class 11 Mathematics
Chapter 7 – Permutations and Combinations
Fundamental principles of counting:
● Fundamental principle of multiplication – If there are three different events
such that one event occurs in
m
different ways, second event happens in
n
different ways and the third event occurs in
p
different ways, then all three
events simultaneously will happen in
m n p
ways.
● Fundamental principle of addition – If there are two jobs such that first wone
can be performed independently in
m
number of ways and the second work
independently can be done in
n
number of ways, then wither of the two jobs can
be performed in
mn
ways.
Some basic arrangements and selections:
● Permutations – It is the linear arrangements of distinct objects taken some or
all at a time. The number of arrangements possible is called the permutations. If
we have two positive integers
r
and
n
such that
l r n
, then the total number
of arrangements or permutations possible for
n
distinct items taken
r
at a time
is mathematically given by,
n r !
Pn!
r
n
.
● Combinations – If we have to select combinations of items from a given set of
items such that the order or arrangement doesn’t matter, then we use
combinations. Such that to find the number of ways of selecting
r
objects from
a set of
n
objects, then mathematically it is given by,
r! n r !
Cn!
r
n
Geometric applications of
Cr
n
:
● If there are
n
non-parallel and non-concurrent lines, then the number of points
of intersections is given by
C2
n
. Similarly, having
n
points, the number of lines
will be
C2
n
such that no three points are collinear.
● If in
n
number of points, there are
m
collinear, then number of straight lines is
given by
C C 1
22
nm
.

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