1.9 Permutations and
Combinations:
1.9.1
Permutations
Problem :
A team has 10 players. Only six can fit into one photo frame. Manager has to be
in each of the photo session. A photographer was asked to give an estimate for all
the different combinations of photos. A photo with frame costs Rs.22. Find out
the estimate given by photographer to the Manager.
Problem :
In a hexagon how many triangles can be formed?
Is it not
interesting to solve the above problems?
Introduction:
We have learnt that the sum
of 1^{st} ‘n’ natural numbers is given by
_{}= 1+2+3+4
…..+n =_{}
What if we multiply these
natural numbers instead of adding them?
1*2=2
1*2*3 =6
1*2*3*4 = 24…
We use the notation n!(
called ‘n factorial’) to denote the product of ‘n; consecutive natural number from 1.
n!=
1*2*3*4….*n
What do we observe?
1! =1
2!=
1*2=2=2*1!
3!=1*2*3=6
=3*2!
4! =1*2*3*4 = 24= 4*3!
n!
= n*(n1)*(n2)*(n3)*…………3*2*1=n*(n1)!
_{} n!
= n*(n1)! Or n = _{}
1.9.1 Example1 :
Let A , B and C be students of your class . You are
asked to arrange them in rows as follows:
1. with rows of 2 students
2. with rows of 3 students
How many arrangements can
you make in each of the case?
Working:
1.9.1.1: Arrange 2
rows with two
students in each row
1.Let
us ask A to stand first. We are left with B and C. We can ask B or C to stand
behind A ( so we have 2 arrangements AB and AC)
2.Let
us ask B to stand first. We are left with A and C. We can will ask A or C to stand behind B (so
we have 2 arrangements BA and BC)
3.Let
us ask C to stand first. We are left with A and B. We can ask A or B to stand behind C (so we
have 2 arrangements CA and CB)
I position 
A 
B 
C 

II
position 
B 
C 
A 
C 
A 
B 
In all we have 6(=3*2) ways
of arrangements: (AB, AC), (BA, BC), (CA, CB)
1.9.1.2: Arrange 3
rows with 3
students in each row:
1.Let
us ask A to stand first. We are left with B and C. We can ask B or C to stand behind
A( so we have 2 arrangements ABC and ACB)
2.Next
let us ask B to stand first. We are left with A and C. We can ask A or C to
stand behind B(so we have 2 arrangements BAC and BCA)
3.Next
let us ask C to stand first. We are left with B and A. We can ask A or B to
stand behind C (so we have 2 arrangements CAB and CBA)
I position 
A 
B 
C 

II position 
B 
C 
A 
B 
C 
A 
III
position 
C 
B 
C 
C 
B 
C 
In all we have 6(=3*2) ways
of arrangements: (ABC, ACB), (BAC, BCA), (CAB, CBA)
1.9.1 Example 2:
Let us take the case of 4 students A, B, C and D in the above example. How many
different Queues can you form?
1. With Queues of 2
students
2. With Queues of 3
students
How many arrangements can
you make in each of the case?
Working:
1.9.1.2.1: Arrange
Queues of 2 students
I position 
A 
B 
C 
D 

II position 
B 
C 
D 
A 
B 
C 
D 
A 
B 
C 
D 
A 
Totally we have 12(=4*3) ways
of arrangements (AB, AC, AD),( BA, BC, BD),( CA, CB, CD),( DA, DB, DC
1.9.1.2.2: Arrange
Queues of 3 students:
I position 
A 
B 
C 
D 

II position 
B 
C 
D 
A 
B 
C 
D 
A 
B 
C 
D 
A 

III
position 
C 
D 
B 
D 
C 
D 
B 
D 
C 
D 
B 
D 
C 
D 
B 
D 
C 
D 
B 
D 
C 
D 
B 
D 
We have:
6(=3*2) Queues with A in
the front (ABC, ABD, ACB ACD, ADB, ADC )
6 Queues with B in the
front (BAC, BAD, BCA, BCA, BDA, BDC)
6 Queues with C in the
front (CAB, CAD, CBA, CBD, CDA, CDB)
6 Queues with D in the
front (DAB, DAC, DBA, DBC, DCA, DCB)
Totally we have 24(=4*3*2)
ways arrangements
The number of arrangements
(‘permutations’) of ‘n’ things taken ‘r’ things at a time is denoted by _{n}P_{r}
Let us analyse
our workings:
Examples 
Total Number of students(n) 
Number of students in the row(r) 
Number of arrangements 
Notation 
Meaning 
Example
1.1 
3 
2 
6 
_{3}P_{2} 
Permutation
of 3 objects taken 2 at a time 
Example
1.2 
3 
3 
6 
_{3}P_{3} 
Permutation of 3 objects taken 3 at a time_{} 
Example
2.1 
4 
2 
12 
_{4}P_{2} 
Permutation of 4 objects taken 2 at a time_{} 
Example
2.1 
4 
3 
24 
_{4}P_{3} 
Permutation of 4objects taken 3 at a time_{} 
_{ }
Permutation is a way of
arranging objects in an orderly way.
1.9.2
Fundamental Principles of counting:
Assume that you and your
friend go to school together. Assume that there are 4 routes to go to your
friend’s house from your house and 3 routes from your friend’s house to the
school. Also, Assume that you have a pet dog ‘Jony’ which some times
follows you.
Find out how many different
routes, ‘Jony’ can
take to reach your house from the school, via your friend’s house?
Note: There are 4 routes to
reach your friend’s house from your house and there are 3 routes to reach
school from your friend’s house.
‘Jony’ can take any of the three routes
(A or
B or
C)
from the school to reach your friends house. From your friend’ house it can
reach your house in four ways (1 or 2 or 3 or 4).
The following table lists
the different ways that ‘Jony’ can take to reach your house from the school via
your friend’s house.
No 
From School to friend’s house 
From friend’s house to your
house 
Route 
1 
A 
1 
A1 
2 
2 
A2 

3 
3 
A3 

4 
4 
A4 

5 
B 
1 
B1 
6 
2 
B2 

7 
3 
B3 

8 
4 
B4 

9 
C 
1 
C1 
10 
2 
C2 

11 
3 
C3 

12 
4 
C4 
‘Jony’ can chose
12(=3*4) ways of reaching your house from the school.
Similarly there are 12 (=4*3) ways of reaching
school from your house.
In summary, if an
activity can be done in ‘m’ ways and another activity is done in ’n’ ways then
the 2 activities together can be done in (m*n) ways.
To find the formula for number of arrangements (Permutations)
possible among ‘n’ things with ‘r’ things taken at a time.
Method:
Let us assume that we have
‘n’ number of objects and ‘r’ number of boxes.
Let boxes be numbered as 1,2,3,….(r1),r.
Our task is to fill these
boxes by taking ‘r’ objects at a time from ‘n’ number of objects.
Box No. 
1 
2 
3 
…… 
(r1) 
r 
No of
ways 
n 
(n1) 
(n2) 

n(r2) 
n(r1) 
1. First box can be filled in n different ways.
2. Second box can be filled
in (n1) different ways.
3. Third box can be filled in
(n2) different ways.
…
r. r’th
box can be filled in
(nr+1) different ways.
By fundamental principle, r
boxes can be filled in
n*(n1)*(n2)*(n3)…..(nr+1)
ways
This is the number of permutations of ‘n’ things taken ‘r’ at a time and is denoted by _{n}P_{r}_{}
_{} _{n}P_{r} =
n*(n1)*(n2)*(n3)…..(nr+1) =======è(1)
By substituting r=n in the
above equation we get
_{n}P_{n}
= n*(n1)*(n2)*(n3)…..(nn+1)
= n*(n1)*(n2)*(n3)…..*1
_{} _{n}P_{n} =n!
Let us multiply and divide
RHS of equation (1) by the same term (nr)*(nr1)*…..3*2*1 we get
_{n}P_{r}
= {n*(n1)*(n2)*(n3)…..(nr+1)*
(nr)*(nr1)*…..3*2*1}/{(nr)*(nr1)*…..3*2*1}
=_{} {_{}n!= 1*2*3……*n and (nr)! =
1*2*3….*(nr)}
Since _{n}P_{r}= _{},
Note:
_{ n}P_{1}=
_{}
= _{}
= n
_{} _{n}P_{1}
=n
_{n}P_{(}_{n1)}
=_{} (Substitute r = (n1) in _{n}P_{r})
= n! (_{}1!= 1)
_{} _{n}P_{(}_{n1)}=
n!= _{n}P_{n}
(nr)!
= n!/_{ nPr}{_{} _{n}P_{r}=
n!/(nr)! }
By substituting r=n in the
above equation we get
0!
= n!/_{ nPn}
= n!/n!
(_{} _{n}P_{n}=
n! )
=1
_{} 0! =1
We summarise the following:
n = n!/(n1)! 
_{n}P_{n} =n! 
_{n}P_{1} =n 
_{n}P_{(n1)}= n!= _{n}P_{n} 
0! =1 
1.9.2 Problem 1 :
In how many ways the letters of the word COMPUTER can
be arranged? How many of these begin with M?
Solution:
Since the number of letters
in the word = 8, we can form 8!=40320 different words
of 8 letters.
Position

1 
2 
3 
4 
5 
6 
7 
8 
Letters 
M 
Fill
from letters among C,O,P,U,T,E,R 
If M is fixed in the first
place then we are left with (n=7) letters to fill the other 7 places.
Then, the number of possibilities
are = 7! =5040 words
1.9.2 Problem 2:
How many 3 digit numbers can be formed using the digits2,3,4,5and
6 without repetitions? How many of these are even numbers?
Solution:
We know _{n}P_{r}_{
}=_{} : It is given that n=5(2,3,4,5,6)
and r=3
Hundred 
Ten 
Unit 
Choose
From (2,3,4,5,6) 
Number of 3 digit numbers which can be
formed = _{5}P_{3} = _{} = _{}=60
Among these, even numbers
are those ending with 2,4 and 6(unit place is 2 or 4 or
6)
Let us find out how many 3
digit numbers can be formed that ends with 2.
Hundred 
Ten 
Unit 
Choose
From (3,4,5,6) 
2 
Even numbers:
Since the unit place is
already taken by 2,
we can only have 3, 4,5and 6 in hundredth and tenth places.
Since 2 will always be in
the units place, we only
need to find all the 2
digit numbers possible with 3,4,5 and 6(r=2).
The number of 2 digit
numbers that can be formed using (n=4) digits (3,4,5,6)
are =_{4}P_{2}= _{}= 4*3 = 12
So with 2 in units place we
can have 12 3digit numbers.
Similarly with 4 in units
place we can have 12 3digit numbers.
Similarly with 6 in units
place we can have 12 3digit numbers.
In all we can have 36(=12+12+12) even numbers.
1.9.2 Problem 3:
How many 3 digit numbers can be formed using 0,1,2,3?
Solution:
Here n=4 and r=3.
The number of 3 digit
numbers that can be formed are _{4}P_{3} =_{} = 4!=24
However, in case of 3 digit
numbers, a number starting with 0 can not be considered as a 3 digit number
(012 is not a 3 digit number but it is a 2 digit number).
Hence from the result we
need to subtract the number of 3digit numbers starting with 0.
Let us find out the number
of 3digit numbers starting with 0 .So we have n=3( 1,2,3)
and r=2
With 0 as the first digit,
number of 3digit numbers we can make=_{3}P_{2} = 3! =6.
Thus we can make 18(=246) 3digit numbers from 0,1,2,3.
1.9.2 Problem 4:
In how many ways can 7 different books be arranged in a shelf? In how many ways
can we arrange three particular books so that they are always together?
Solution:
Here n=7. So the number of
ways these books can be arranged is 7! = 5040.
Let the books be A,B,C,D,E,F,G.
It is said that three books need to be together. Let they be B, C and D. For easy
understanding we can tie these 3 books together and call this bundle as H. So we are left with A,H,E,F,G(5
objects). These can be arranged in 5!=120 ways.
Since H is bundle of 3
books (B,C,D). The books in the bundle H themselves
can be arranged in 3!=6
ways.
Thus we have 6*120=720
ways of arranging 7 books such that 3 books are always together.
1.9.3
Combinations:
1.9.3 Example 1 :
Let A , B and C be students of your class. A photographer was asked to take
photos such that:
1. Different combinations
of 2 students in one snap shot
2. Different combinations
of 3 students in a snap shot
How many shots does he need
to take?
Working:
Example 1.1: Photo
session for group of 2 students
It can be seen from the
example 1.9.1.1.1 that,
the following arrangements are possible
I position 
A 
B 
C 

II position 
B 
C 
A 
B 
C 
A 
But for a photo session we
notice that AB = BA, BC = CB, and CA=AC.
Though there are 6
arrangements possible, we only have 3 unique combinations (AB, BC, CA).
Example 1.2: Photo
session for group of 3 students
It can be seen from the
example 1.9.1.1.2 that, the following arrangements are possible
I position 
A 
B 
C 

II position 
B 
C 
A 
B 
C 
A 
III position 
C 
B 
C 
C 
B 
C 
But for a photo session all
the 6 combinations are same.
Though there are 6 arrangements
possible, there is only one unique combination (ABC).
1.9.3 Example 2:
Let A B C D be 4 students in your class . A
photographer was asked to take photos such that:
1. Different combinations
of 2 students in one snap shot
2. Different combinations
of 3 students in one snap shot
How many snap shots does he
need to take?
Working:
Example 2.1: It can be
seen from the example 1.9.1.2.1 that, the following 12 arrangements are possible
I position 
A 
B 
C 
D 

II position 
B 
C 
D 
A 
C 
D 
A 
B 
D 
A 
B 
C 
Notice that for a photo
session AB=BA, AC=CA, AD=DA, BC=CB, BD=DB and CD=DC.
Though 12 arrangements are possible , we only
have 6(=3*2) unique combinations (AB, AC, AD, BC, BD, CD).
Example 2.2: It can be
seen from the example 1.9.1.2.2 that, the following 24 arrangements
are possible
I position 
A 
B 
C 
D 

II position 
B 
C 
D 
A 
C 
D 
A 
B 
D 
A 
B 
C 

III position 
C 
D 
B 
D 
B 
C 
C 
D 
A 
D 
A 
C 
B 
D 
A 
D 
A 
B 
B 
C 
A 
C 
A 
B 
Note for a photo session
ABC=BAC=ACB=BCA=CAB=CBA
ABD=ADB=BAD=DAB=DBA=BDA
ACD=ADC=CAD=DAC=DCA=CDA
BCD=BDC=CBD=CDB=DBC=DCB
are
all same.
Though 24 combinations are
possible, there are only 4
unique combinations. (ABC, ABD, ACD, BCD)
The number of combinations
of ’n’ things taken ‘r’ things at a time is denoted by _{n}C_{r}
Let us analyse
our results
Examples 
Total Number of students(n) 
Number of students Per shot 
Number of combinations 
Notation 
Example
1.1 
3 
2 
3 
_{3}C_{2} 
Example
1.2 
3 
3 
1 
_{3}C_{3} 
Example
2.1 
4 
2 
6 
_{4}C_{2} 
Example
2.2 
4 
3 
4 
_{4}C_{3} 
Let us find the
relationship between permutations(1.9.1) and combinations(1.9.3) using
the examples 1 and 2 of Permutations and Combinations.
Examples 
Number of students(n) 
Number of students taken at a time(r) 
Number of Permutations_{ nPr}) (1.9.1)_{} 
Number of _{} Combinations(_{n}C_{r}) (1.9.3) 
_{n}P_{r}/_{n}C_{r}_{} = 
Example
1.1 
3 
2 
6= _{3}P_{2} 
3=_{3}C_{2} 
2=2! 
Example
1.2 
3 
3 
6= _{3}P_{3} 
1=_{3}C_{3} 
6=3! 
Example
2.1 
4 
2 
12= _{4}P_{2} 
6=_{4}C_{2} 
2=2! 
Example
2.2 
4 
3 
24= _{4}P_{3} 
4=_{4}C_{3} 
6=3! 
We notice that _{n}P_{r}= _{n}C_{r}_{ }*
r! _{n}C_{r}_{ }=_{
nPr÷r}!
Formula for number of combinations of ‘n’ things taken
‘r’ at a time:
In the example 2.1
discussed above we notice the following
(Permutations of ‘n’ things taken ‘r’ at a time)= (Selection (Combination) of ‘n’
things taken ‘r’ at a time)*(Arrangement of ‘r’ things)
_{n}P_{r}_{ }= _{n}C_{r}*_{ rPr}
1.9.3 Problem 1:
If _{n}P_{r}
= 336 and _{n}C_{r}=56 find n and r
Solution:
We know _{n}P_{r}÷_{n}C_{r}_{
}= r!
_{}r!= _{} = 6=3*2*1=3!
_{} r=3
Substitute r=3 in _{n}C_{r}=_{ nPr÷r}!
= {n! ÷ (nr)} ÷r!
= {n*(n1)*(n2)*(n3)! ÷
(n3)! }÷3!
_{}56 = n*(n1)*(n2) ÷6
I.e. 56*6 =336 =
n*(n1)*(n2) = 8*7*6
_{} n=8
1.9.3 Problem 1:
A king has 8 different types of ornamental jars in his courtyard. Tell me the
number of different combinations in which theses can be arranged?
(Lilavati Shloka 116)
Solution:
Total number of jars(n) =8
No 
Type
of arrangement 
Nos 
1 
No of
arrangements with 1 jar at a time 
_{8}C_{1} 
2 
No of arrangements with 2 jars at a time 
_{8}C_{2} 
3 
No of arrangements with 3 jars at a time 
_{8}C_{3} 
4,5,6 
 

7 
No of arrangements with 7 jars at a time 
_{8}C_{7} 
8 
No of arrangements with 8 jars at a time 
_{8}C_{8} 
Total number of
arrangements = _{8}C_{1}+
_{8}C_{2}
+ . . . + _{8}C_{7}
+ _{8}C_{8} =255 = 2^{8}1
1.9.3 Problem 3 :
A Marriage bureau is in the business of identifying girls
for boys and boys for girls. They are having 5 girls and 4 boys in their
register looking for a match. In how many ways they can make proposals with 2
boys and 2 girls?
Solution:
1. There are 4 boys. The
number of ways in which 2 boys can be grouped are_{ 4}C_{2}=4*3*2!/2!*2!=6
2. There are 5 girls. The
number of ways in which 2 girls can be grouped are _{5}C_{2}=5*4*3!/3!*2! = 10
For every group of 2 boys
selected out of 6 groups, we can match with every group of 2 girls selected out
of 10 groups.
_{} Total number of
proposals possible = 6*10=60
1.9.3 Problem 4 : A team has 10 players. Only
six can fit into one photo frame. Manager has to be in each of the photo
session. A photographer was asked to give an estimate for all the different
combinations of photos. A photo with frame costs Rs.22. Find out the estimate
given by photographer to the Manager
Solution:
Here n =10, r=5( manager has to be in every photo)
The number
of 5member groups teams that can be formed from 10 members are
= _{10}C_{5}
= 10!/5!*5!
= (10*9*8*7*6*5!)/(5!*5!)
= 10*9*8*7*6/120(_{} 5! gets cancelled)
= 9*4*7 =252 photos
Total cost = 252*22= Rs.5, 544
1.9.3 Problem 5 :
Your school has one teacher for each of the subjects :
Mathematics, Social science, General Science, Moral science, English, Hindi,
Local Language, Physical Training. One of them is a headmaster.
(a) How many committees of
5 members can be formed?
(b) How many of them will
not have Headmaster in them?
Solution:
Number of teachers (n) = 8
Number of teachers in the
committee (r) =5
_{} Number of committees
that can be formed is
= _{8}C_{5}=
8!/(85)!*5!= 8*7*6*5!/3!*5!= 8*7*6/6 = 56
If we are to have Headmaster in the committee,
then we are left with only 7 teachers for selection. In such a case
The number of teachers (n)
= 7.
Since headmaster is already
a member of the committee, we only need to form a committee of 4 members (r) =4.
_{} Number of committees
with head master is
= _{7}C_{4}= 7!/(74)!*4!= 7*6*5*4!/3!*4!= 7*6*5/6 = 35
No of committees without
headmaster = total number of committees – Number of committees with head master
= 5635 =21
1.9.3 Problem 6 :
There are 20 non collinear points on a plane. How many (a) straight lines (b)
triangles can be formed by joining these points?
Solution:
The
number of non collinear points (n=20) Since
straight line has 2 points (r=2), Total
number of straight lines that can be formed are = _{20}C_{2}=
20!/(202)!*2!= 20*19*18!/18!*2! =
20*19/2 = 190 Since
triangles are formed with 3
points(r=3), Total
number of triangles that can be formed are = _{20}C_{3}=
20!/(203)!*3!= 20*19*18*17!/17!*3! =
20*19*18/6 = 1140 

Solution:
1.9.3 Problem 6 :
In a hexagon how many triangles can be formed?
Solution:
The number of vertices in a
hexagon (n)=6
The number of points
required for a triangle(r) =3
_{} Number
of triangles that can be formed in a hexagon are
= _{6}C_{3}=
6!/(63)!*3!= 6*5*4*3!/3!*3!= 20
1.9.3 Problem 7: A country had selected a
team of 8 batsmen and 7 bowlers to play a cricket match in
Solution:
If he chooses to have 5
bowlers then he can only have 6 batsmen out of 8.
If he chooses to have 6
bowlers then he can have 5 batsmen out of 8.
For every group of bowlers the
captain chooses, he has several choices of groups of batsmen and hence total
choice available for a particular option is product of these two choices.
Possibilities 
Total bowlers available =7 
Total batsmen available =8 
No of combinations 

1 
5 (r=5) 
6 (r=6) 
=_{7}C_{5}*_{8}C_{6} 
21*28=588 
2 
6 (r=6) 
5 (r=5) 
=_{7}C_{6}*_{8}C_{5} 
7*
56=392 
The captain has a choice of 980(588+392)
choices!!!
1.9.3 Problem 8: Indian
parliament has selected 8 members to form a 5member Social Welfare Committee.
The Health Minister and Women Welfare Ministers are among these 8 members. The
committee’s role is to give recommendations for improvement of Welfare of
Citizens. There is a rule that the committee should be headed by a Minister and
there can not be more than one minister in the same committee. Find out how
many committees can be formed as per the rule?
Solution:
Since the committee has to
be headed by a minister, one of the ministers has to be in the committee. With
one minister being a member of committee and two ministers can not be in the
same committee,
Number of members available
for selection is (n=6)(=82)
Since one minister is
always in the committee, we need to select only 4 members for the committee
(r=4)(=51)
_{} The no of ways
committee can be formed =_{6}C_{4}= 15
1.9
Summary of learning
Properties 
Permutations 
Combinations 
Meaning ====> 
Arrangement
of things in an orderly manner 
Selection
of different objects 
Example ====> 
How
many different words can be formed from the letters of the word ‘MATHS’ 
How
many 10 member hockey team be
formed from a list of 20 players 
Definition ====> 
No of ways
of arranging ‘n’ things taken ‘r’ things at a time 
No of
combinations of ‘n things taken ‘r’ things at a time 
Formula ====> 
_{n}P_{r} = n(n1)(n2)…(nr+1)= _{} 
_{n}C_{r}_{ }= _{n}P_{r} /r! 
Relationship ===> 
_{n}P_{r}= _{n}C_{r}_{ }* r! _{} 
_{ } 