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!/_{ n}P_{r}{_{} _{n}P_{r}= n!/(nr)! }
By substituting r=n in the above equation we get
0!
= n!/_{ n}P_{n}
= 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_{ n}P_{r}) (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 }=_{ n}P_{r}÷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}*_{ r}P_{r}
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}=_{ n}P_{r}÷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! _{} 
_{ } 