1. science
2. mathematics
3. arithmetic

4
Fundamental principle of
counting
Permutation
Factorial notation
n
Pr
n!
n r !
Combination
n
Cr
n!
r! n r !
Permutations and
Combinations
This unit facilitates you in,
stating the fundamental principle of
counting.
defining permutation and combination.
using factorial notation.
deriving the formula for numbe r of
permutations (nPr).
deriving the formula for numbe r of
combinations (nCr).
differentiating between permutations and
combinations
solving problems using nPr and nCr.
Jacob Bernoulli
(1654 - 1705 A. D.)
Swiss mathematician, Jacob
Bernoulli has given a complete
treatment of permutations
and combinations in his book
‘Ars Conjectandli’ (Posthumously
published in 1713 A. D)
In mathematics the art of proposing a question
must be held of higher value than solving it.
- George Cantor
66
UNIT-4
Let us begin this unit with some real life situations. Study them and discuss in groups.
Illustration 1
We see different types of motor vechicles plying on the road. Each and every vehicle will
have a number plate with it's registration number.
Recall that each number plate will have the following information.
1) Name of the state (Eg : KA - for Karnataka)
2) Regional transport office code number.
(05 - Bangalore - Jayanagar RTO)
EV4007
3) English alphabet/s
4) Followed by a number (from a single digit to a four digit number)
Do you know how many vehicles can be registered in any one of the series? You
know that there can be 9999 vehicles!!
Have you ever thought how these numbers are generated by the R.T.O? What
mathematical principle is behind this calculation?
Illustration 2
Suppose we have a suitcase with number lock. The number lock has three wheels each
labelled with ten digits from 0 to 9. The lock can be opened if three specific digits are
arranged in a particular arrangement or may be in a sequence say with no repetitions.
Let us suppose that we have forgotten this sequence of digits. In order to open the lock,
how many sequences of three - digits we have to check? To answer this question we have
to check all the three - digit numbers. But this will take lot of time. Is there any
mathematical idea which can help us to known the number of possibilities?
In this unit let us learn some basic counting techniques by which we are able to
find solutions for the above situations.
As a first step, we shall examine carefully a principle which is most fundamental to
learning of the counting techniques.The main subject of this unit is counting. Given a
set of objects, the task is to arrange a subset according to some specifications or to select
a subset as per some specifications.
Fundamental Principle of Counting (FPC)
Study the following illustrations.
Example 1: There are three dolls as shown.
Two out of them have to be arranged on a shelf. In how many different ways can they be
arranged? How do we count this?
Observe the tree diagram representing the number of ways of selecting the first
doll and the second doll.
S ––– PS1
2
Permutations and Combinations
67
3
4
From this diagram, we can conclude that the first doll can be
1
selected in three different ways and for each of these first selection,
2
the second doll can be selected in two different ways.
3
So there are totally 6 ways.
4
The total number of ways of arranging two dolls out of three
is 3 × 2 = 6 ways.
The 6 different ways can be represented as follows:
Example 2: A boy has 2 pants and 4 shirts. How many different pairs of a pant and a shirt
can he dress up with?
There are two ways in which a pant can be chosen and for every choice of a pant
there are four ways in which a shirt can be chosen.
Let us denote the two pants as P1 and P2 and the shirts as S1, S2, S3 and S4. The
different ways of pairing a pant and a shirt can be represented by a tree diagram as shown
below.
Shirts
1
P1
Pants
P2
 there are totally 8 ways.
1
S2––– PS
1
S3––– PS
1
S4––– PS
1
S1––– PS
2
S2––– PS
2
S3––– PS
2
S4––– PS
2
In other words, there are 2 × 4 = 8 ways of pairing a pant and a shirt.
Example 3: Sanjay goes to a hotel to have breakfast.
The menu card indicates the following items that are served:
Tiffin
Sweets
Hot drinks
Idli
Kesari bath
Coffee
Dosa
Jamoon
Tea
Poori
Badam milk
Kharabath
In how many different ways can he select one item from each type?
Tiffin can be chosen in 4 different ways. After a tiffin is chosen, a sweet can be
chosen in 2 different ways.
 there are 4 × 2 = 8 ways in which a tiffin and a sweet can be chosen.
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UNIT-4
For each of the above 8 ways, a hot drink can be chosen in 3 different ways.
 there are 8 × 3 = 24 ways in which Sanjay can choose the three items.
 the total number of ways = 4 × 2 × 3 = 24
If we represent tiffin as T1, T2, T3 and T4, sweets as S1 and S2, hot drinks as H1, H2
and H3, the 24 possible ways can be listed in the table as follows.
T1 S1 H1
T2 S1 H1
T3 S1 H1
T4 S1 H1
T1 S2 H1
T2 S2 H1
T3 S2 H1
T4 S2 H1
T1 S1 H2
T2 S1 H2
T3 S1 H2
T4 S1 H2
T1 S2 H2
T2 S2 H2
T3 S2 H2
T4 S2 H2
T1 S1 H3
T2 S1 H3
T3 S1 H3
T4 S1 H3
T1 S2 H3
T2 S2 H3
T3 S2 H3
T4 S2 H3
The data from the above three examples is entered in the table. Study the table and
discuss in groups to infer the principle of counting.
Example
Number of ways
Activity 1
Activity 2
Total number
Activity 3
of ways
1
3
2
–
3×2=6
2
4
2
–
4×2=8
3
4
2
3
4 × 2 × 3 = 24
We can conclude and state the fundamental principle of counting (FPC) or
multiplication principle as follows :
If one activity can be done in 'm' number of different ways and corresponding to each of
these ways of the first activity, second activity (independent of first activity) can be
done in 'n' number of different ways then, both the activities, one after the other can be
done in (m × n) number of ways.
The above principle can be generalised for any finite number of events.
For three activities, the FPC can be stated as:
If one activity, can be done in 'm' number of different ways, for each of these 'm'
different ways, a second activity can be done in 'n' number of different ways and for
each of these activities, a third activity can be done in 'p' ways, then all the three
activities one after the other can be done in (m × n × p) number of ways.
We can now easily find the total number of ways of doing more than two activities
without actually listing, counting or drawing tree diagram. We shall illustrate this FPC
with a few more examples.
ILLUSTRATIVE EXAMPLES
Example 1: How many 2 - digit numbers can be formed from the digits {1, 2, 3, 4, 5}
without repetition and with repetition?
Sol: The digits in the selection set are {1, 2, 3, 4, 5}.
U
T
Since we have to form 2 - digit number, let us draw 2 boxes, one for
units place and another for tens place.
Permutations and Combinations
69
We fill the units place first and then tens place.
The box for units place can be filled in 5 different ways with the digits 1, 2, 3, 4
and 5.
After filling the units place we are left with 4 digits, as repetition of digits are not
allowed, the box for tens place can be filled in 4 different ways.
By applying fundamental principle of counting we get 5 × 4 = 20 ways.
 there are 20 ways of forming 2 - digit numbers by using 1, 2, 3, 4, 5 without
repetition.
If the repetition of digits are allowed then the tens place can also be filled by 5
different ways, using 1, 2, 3, 4 and 5.
 The total number ways a 2 - digit number can be formed by using 1, 2, 3, 4 and 5
with repetition is 5 × 5 = 25 ways.
Note: Discuss in groups and verify the answer by actually listing the 2 - digit numbers.
Example 2. How many 3 letter code can be formed by using the five vowels without
repetitions?
Sol. The vowels are a, e, i, o and u.
We have to form a three letter code.
The first letter can be chosen in 5 ways.
The second letter can be chosen in 4 ways.
The third letter can be chosen in 3 ways.
( no repetition is allowed)
( no repetition is allowed)
The total number of ways in which the 3 letter code can be formed without repetition
= 5 × 4 × 3 = 60 ways
Example 3. How many 3 - digit numbers can be formed from the digits 0, 1, 2, 3 and 4 with
repetitions?
Sol. A Three - digit number will have three places - Hundreds, tens, units.
The digits in the selection set are {0, 1, 2, 3, 4}
Hundreds place - Zero cannot occupy hundreds place, because it becomes a 2 - digit
number. Hence, the hundreds place can be filled in by 4 ways, i.e. 1, 2, 3, 4.
Tens place - We have five ways of filling tens place, as repetitions are allowed and
zero can also be used.
Units place - This can also be filled in 5 different ways.
 the total number of 3 - digit numbers that can be formed usi ng
0, 1, 2, 3, 4 are 4 × 5 × 5 = 100 numbers.
Example 4. Now let us try to solve the suitcase lock
problem (illustration 2 in page 2), by applying the
fundamental principle of counting.
Sol. The number lock of the suitcase has three digits.
Each number can be formed in 10 different ways
using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
6 2 0
7 3 1
8 4 2
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UNIT-4
It is also evident that the digits can be repeated.
 the total number of ways the number lock code can be formed is 10 × 10 × 10 =
1000 ways
So, to open the suitcase 1000 different ways have to be checked.
EXERCISE 4.1
1. How many 3 - digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 without
repeating any digit.
2. How many 3 digit even numbers can be formed using the digits 3, 5, 7, 8, 9, if the
digits are not repeated?
3. How many 3 letter code can be formed using the first 10 letters of English alphabet,
if no letter can be repeated?
4. How many 5 digit telephone numbers can be formed using the digits 0 to 9, if each
number starts with 65 and no digit appear more than once?
5. If a coin is tossed 3 times, find the number of outcomes.
6. Given 5 flags of different colours, how many different signals can be generated if
each signal requires the use of 2 flags one below the other?
Permutations and Combinations
While learning the fundamental principle of counting, we have observed that in all the
examples some number of objects are given and out of these a few objects are to be chosen.
Now, let us study more about the selection and arrangement of things.
Study the examples given in the table.
Read the first problem in column 'P' and then the first problem in column 'C' Now compare
them. Repeat the same procedure for the remaining set of problems.
Sl No.
Column 'P'
Column 'C'
3 persons contest for the post of a
president and a secretary. In how
many ways they can be elected?
There are 3 condidates contesting for
2 posts. In how many ways can the
posts be filled.
2.
A child has 8 frocks and 4 pairs of
shoes. In how many ways the child
can dress herself?
There are 8 white and 4 red roses in
a garden. In how many ways 4 flowers
of which 2 are red be plucked?
3.
In how many ways can all the
letters of the word TEACH be
arranged?
In how many ways all the letters of
the word MEANS can be selected?
1.
Do you notice any similarties in the problems of each column? For the problems in the
column 'P', we have to find out the number of ways of arrangements. For the problems in
the column 'C' we have to find the number of ways of selections.
Permutations and Combinations
71
In solving prolems of the above types the first step is we must carefully identify,
whether it is the arrangements or selections. As an example let us understand the first
problem in the above table.
Let the 3 persons be A, B and C. There
are two posts, president and a secretary.
Let us have the box notation and write all
the possible ways.
President
Secretary
1
2
3
4
A
A
B
C
B
B
A
C
5
6
C
A
C
B
Case 1. If 'A' is the president then 'B' can
be secretary or 'C' can be secretary
Case 2 : If 'B' is the president then 'A' can
be secretary or 'C' can be secretary
Case 3: If 'C' is the president then 'A' can
be secretary or B can be the secretary.
So all together there 6 possible ways of
electing.
Let the 3 persons be A, B and C. There are
two posts only. Let us have the box notation
and write all the possibilities.
Choices for posts
SL
No.
Post - 1
1
A
B
2
A
C
3
B
C
Post - 2
Case 1. If 'A' is selected for one post,
another post can be given to B or C.
Case 2 : If 'B' is selected for one post then
the other post can be given to C but not to
A because BA and AB are the same
selections. Hence there can be 3 ways by
which the posts can be filled up
It is also evident from the above discussion that, in examples of column P the
arrangement of things or objects or persons is done with regard to order. In examples of
column C, the selection of things or objects or persons is done without regard to order.
Such arrangements with regard to order are called permutations and selections
without regard to order are called combinations.
Permutation : A permutation is an ordered arrangement of a set of objects.
It is an act of arrangements of objects in an orderly way.
A permutation of n elements taken r at a time is any ordered subset of r elements
taken from the set of n elements.
The number of permutations of 'n' different objects taken 'r' at a time is denoted as
Pr where r  n.
n
Combination : A combination is a selection of a set of objects without any order. It is an
act of selection of objects not involving any orderly way.
A combination of n elements taken r at a time is any subset of r elements taken from
the set of n elements (without regard to order).
Note that in a set there is no order for listing the elements. e.g. {1, 3, 2} = {2, 1, 3}
The number of combinations of 'n' different objects taken 'r' at a time is denoted as
Cr, where r  n.
n
72
UNIT-4
EXERCISE 4.2
I. Below are given situations for arrangements and selections. Classify them as
examples of permutations and combinations.
1. A committee of 5 members to be chosen from a group of 12 people.
2. Five different subject books to be arranged on a shelf.
3. There are 8 chairs and 8 people to occupy them.
4. In a committee of 7 persons, a chairperson, a secretary and a treasurer are to be
chosen.
5. The owner of children's clothing shop has 10 designs of frocks and 3 of them have to
be displayed in the front window.
6. Three-letter words to be formed from the letters in the word 'ARITHMETIC'.
7. In a question paper having 12 questions, students must answer the first 2 questions
but may select any 8 of the remaining ones.
8. A jar contains 5 black and 7 white balls. 3 balls to be drawn in such a way that 2 are
black and 1 is white.
9. Three-digit numbers are to be formed from the digits 1, 3, 5, 7, 9 where repetitions
are not allowed.
10. Five keys are to be arranged in a circular key ring.
11. There are 7 points in a plane and no 3 of the points are collinear. Triangles are to
be drawn by joining three non-collinear points.
12. A collection of 10 toys are to be divided equally between two children.
In each of the above examples give reason to explain why it is permutation or
combination.
Try: List atleast 5 examples each for permutation and combination. Discuss in the
class.
To find the general formula for the number of ways by which rth box can be filled up:
By using FPC we can write the expansion of each permutation as follows :5
P5
5 4 3 2 1
5
P4
5 4 3 2
5
P3
5 4 3
5
P2
5 4
5
P1
5
In each permutation, observe
carefully the number of ways in
which the last box is filled up.
There is some common pattern in all
the permutations.
There is a relationship between
values of n, r and the number of
ways in which the last box can be filled up.
The relation is as follows :
Permutations and Combinations
73
'Subract 'r' from 'n' and add one to it' to get the number of ways by which rth box can be filled
up i.e., (n – r + 1)
Note that when use say rth box, r is atleast 1. So r > 0
Study the following examples:
1) Consider 5P5 - we know that the last box i.e., 5th box can be filled up in just 1 way. We
can get this by subtracting (5 from 5) and add 1 to it.
i.e., (5 – 5 + 1) = 0 + 1 = 1 way.
2) Consider 5P4 - Here also we know that the last box i.e., 4th box can be filled up in 2
ways. We can get this by subtracting (4 from 5) and add 1 to it.
i.e., (5 – 4 + 1) = 1 + 1 = 2 ways.
3) Consider 5P3 - Here also we see from the above table last box can be filled up in 3
ways. This we can get by subtracting (3 from 5) and then add 1 to it.
i.e., (5 – 3 + 1) = 2 + 1 = 3 ways
Hence, we can conclude that in nPr the rth box can be filled in (n – r + 1) ways.
To find the number of permutations of 'n' distinct objects taken 'r' at a time where
0  r  n and the objects do not repeat.
The number of permutations of 'n' objects taken 'r' at a time is same as the no of ways of
filling 'r' blank boxes with 'n' given objects.
Consider 'n' different objects and 'r' blank boxes as shown below.
1
2
3
n (n 1) (n 2)
ways ways ways
r –1 r
[n
r – 2)] [n r –1)]
ways
ways
The first box can be filled by 'n' number of ways by putting 'n' different given objects.
Thus, there are 'n' different ways of filling up the first box.
After filling up the first box by any one of the 'n' objects, we are left with (n – 1)
objects as repetitions are not allowed.
So, the second box can be filled up in (n – 1) number of ways.
Thus, the first 2 boxes can be filled up in n (n – 1) number of ways (by FPC).
Now after filling up the first 2 boxes, we are left with (n – 2) objects.
Thus, the 3rd box can be filled up in (n – 2) number of ways.
Again by F.P.C the first three boxes can be filled up in n(n – 1) (n – 2) ways. Observe
that a new factor is introduced with each new box filled up, and that at any stage the
number of factors is the same as the number of boxes filled up. Hence, by FPC the number
of ways of filling 'r' boxes in succession is given as follows.
n  (n – 1)  (n – 2)  ............................. [n – (r – 1)]
74
UNIT-4
Thus the number of permutations of 'n' objects taken 'r' at a time is given by nPr
n

Pr = n  (n – 1)  (n – 2)  ............................. [n – (r – 1)]
Corollary : The number of permutations of 'n' objects taken all at a time is
or
n
Pn = n  (n – 1)  (n – 2)  ............................. [n – (n – 1)]
n
Pn = n  (n – 1)  (n – 2)  .............................

1
n
Pn = n  (n – 1)  (n – 2)  .............................

321
('n' factors)
Factorial notation :
Observe that the RHS of the above expresssion is the product of first n natural numbers.
It is denoted by a notation n!. n! is read as n factorial.
n  (n – 1)  (n – 2)  ............................. 3 2 1 = n!
n! denotes the product of first n natural numbers.
Thus 4! = 1 × 2 × 3 × 4
or
4×3×2×1
3! = 1 × 2 × 3
or
3×2×1
2! = 1 × 2
or
2×1
n! = 1 × 2 × 3 ×..... × n
or
n × (n – 1) (n – 2) × .... × 3 × 2 × 1
0! by definition is taken as 1.
[Note : If n is a negative number or a decimal, n! is not defined.]
Remember :
n! is the product of first 'n' natural numbers.
Sn is the sum of first 'n' natural numbers.
n! = n (n – 1) (n – 2) × ...... × 3 × 2 × 1
Sn = 1 + 2 + 3 + ........... + n
Study the expansions of factorial notations given below.
n! = n (n – 1) (n – 2) × ...... × 3 × 2 × 1
(n – r)! = (n – r) (n – r – 1) (n – r – 2) × ...... × 3 × 2 × 1
(n – r + 1)! = (n – r + 1) (n – r) (n – r – 1) × ...... × 3 × 2 × 1
(n – r – 1)! = (n – r – 1) (n – r – 2) (n – r – 3) × ...... × 3 × 2 × 1
(r – 1)! = (r – 1) (r – 2) (r – 3) × ...... × 3 × 2 × 1
In general ;
n! = n (n – 1) (n – 2) × ...... × 3 × 2 × 1
n! = n [(n – 1) (n – 2) × ...... × 3 × 2 × 1]
n! = n [(n – 1)!]
or
n! = n (n – 1) (n – 2)!
or
n! = n (n – 1) (n – 2) (n – 3)! and so on.
For example, Consider 5!
5! = 5 × 4 × 3 × 2 × 1
5! = 5 × (4!)
5! = 5 × 4 × (3 !)
5! = 5 × 4 × 3 × (2!)
Permutations and Combinations
75
EXERCISE 4.3
1. Convert the following products into factorials.
(i)1 × 2 × 3 × 4 × 5 × 6 × 7 (ii) 18 × 17 × ........ × 3 × 2 × 1
(iii)
6×7×8×9
2. Evaluate :
(iv) 2 × 4 × 6 × 8
(i) 6!
3. Evaluate :
(iv)
7!
5!
(i)
n
(ii) 9!
(v)
n!
r !
(iii) 8! – 5!
12!
9! 3!
(ii) n
(vi)
30!
28!
n!
r !r ! when n = 15 and r = 2
4. Find the LCM of 4!, 5!, 6!
5. (i) If (n +1)! = 12 (n – 1)! find the value of n
(ii) If
1
9!
n
find the value of n
11!
1
10!
6. Simplify : (n
(n 1)!
2)! (n–1)
To derive the formula for nPr in factorial notation
By Fundamental Principle of counting, nPr = n (n – 1) (n – 2) × .......... × (n – r + 1)
Consider the RHS of the equation,
n (n – 1) (n – 2) ........ (n – r + 1)
The first factor of the RHS is n. The last factor of the RHS is (n – r + 1).
 RHS is the product of natural numbers starting from 'n' in descending order upto
(n – r + 1). Had the product continued upto 1 what we would have got? We would have got n!
What extra factor we have to write to get n! ?
[(n – r) (n – r – 1) × ...... × 3 × 2 × 1] which is (n – r)!
n
Pr = n (n – 1) (n – 2) × ..... × (n – r + 1)
n
Pr =
n
Pr =
n (n 1) (n
n!
n r!
2) .... (n r 1) (n r) (n r 1) .... 3 2 1
(n r) (n r 1) ....3 2 1
 nPr =
n!
n r!
[If r > 0]
n
 Number of permutations of 'n' things taken 'r' at a time is Pr
n
n!
r ! , where 0  r  n.
The above formula for nPr holds good only when repetitions are not allowed.
76
UNIT-4
The number of permutations of 'n' different objects taken 'r' at a time, where repetition
is allowed is nr.
Now let us study some special cases of nPr
Case 1: What happens when r = 0? We shall see it now.
If r = 0 in how many ways can we arrange 0 objects from n objects?
Arranging no object at all is the same as leaving behind all the objects as they are
n
1
and we know that there is only one way of doing it. Thus P0
n
n!
n!
=1
Po = n 0 ! =
n!
Example :
100
Case 2: Let

Also,
Example :
Case 3: Let
250
1,
P0
958
1,
P0
1
P0
r=n
n
Pr = n (n – 1) (n – 2 ) × .................... × (n – r + 1)
n
Pn = n (n – 1) (n – 2 ) × .................... × (n – n + 1)
n
Pn = n (n – 1) (n – 2 ) × .................... × 1
n
Pn = n (n – 1) (n – 2 ) × .................... × 3 × 2 × 1
n
Pn = n!
n
n!
n n !
Pn
1009
n!
0!
n!
1
n!
10
P1009 = 1009 !,
(recall the definition 0! = 1)
1400
P10 = 10 !,
P1400 = 1400 !
r=1
n
Pr
n!
n r !
Example :100P1 = 100,
n
 P1
457
n!
n 1!
n
P1
n n
n
1!
1!
 nP1 = n
P1 = 457
Discuss: 5P6 is meaningless. Why?
n
Remember:
If r = 0 then
n
P0 = 1
Pr
n!
n r !
If r = n then nPn = n!
If r = 1 then
n
P1 = n
Permutations and Combinations
77
ILLUSTRATIVE EXAMPLES
(i) 7P3
Example 1. Evaluate :
7!
7 3 !
Sol. (i) 7 P3
8
(ii)
7!
4!
8!
8 5 !
P5
7
8!
3!
6 5
4!
8
7
Example 2. Find 'r ', if 5 4Pr = 6
Sol. 5
4
4!
r !
6
6
5!
5!
4
6
r !
5
(ii) 8P5
5
4!
6 5
3!
7
6 5
4
3!
7
6
5
4
 r2 – 11r +24 = 0
r
4
r !
 (r – 8) (r – 3) = 0
[r = 8 has no meaning]
Example 3. Prove that n! + (n + 1)! = n! (n + 2)
Sol. LHS: n! + (n + 1)! = n! + (n + 1) n!
n
Example 4. Find 'n' if
n (n
(n
1) n
1) n
P4
P4
2 n
2 n
= n! (1 + n + 1)
= n! (n + 2) = RHS
5
3
n 1
3
3 n
5
,
3
4
n
n
4
5
3
5(n – 4) = 3n
 5n – 20 = 3n
 2n = 20
 n = 10
Example 5. If 2n + 1Pn – 1 : 2n – 1Pn = 3 : 5 find n.
2n 1
Sol.
Pn 1
Pn
3
5
2n 1
i.e., 5 ×
2n + 1
5. 2n
2n
1
Pn – 1 = 3 ×
1!
n
1 !
n
2!
10 n 2n
n
2 n
2n – 1
Pn
3. 2n 1 !
2n 1 n !
1 !
5 2n 1 2n 2n
3 2n 1 !
n
1
1 n n
1!
3
1 !
6720
6
6 r 5
r !
 r=3
Sol.
4
Pr – 1
1
(6 – r) (5 – r) = 6
8
5!
r 1 !

r !
210
n
1 !
78
UNIT-4
 10 (2n + 1) = 3(n + 2) (n + 1)
3n2 – 11n – 4 = 0  (3n + 1) (n – 4) = 0
n
1
3
or n = 4.
Since n is a positive integer, n = 4.
Example 6. If nPn = 5040, find 'n'
7 5040
6 720
Sol. nPn = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
n! = 7!  n = 7
5 120
4 24
3 6
n
(ii) If P2 = 90 find 'n'
Sol. nP2 = 90
n(n – 1) = 10 × 9
2
1
2
 n = 10
(iii) If 11Pr = 990 find 'r'
Sol.
11
Pr = 990
11Pr = 11 × 10 × 9
 r=3
EXERCISE 4.4
1. Evaluate:
i)
2.
12
P4
ii)
75
P2
iii) 8P8
1. If nP4 = 20 nP2 find 'n'
2. If 16 nP3 = 13 n+1P3 find 'n'
3. If 5Pr = 2 .6Pr –1 find 'r'
3.
1. If nP4 : nP5 = 1 : 2 find 'n'
2. If n–1P3 : n+1P3 = 5 : 12 find 'n'
4.
1. If 9P5 + 5 9P4 =
10
Pr, find 'r'
2. If 10Pr+1 : 11Pr = 30 : 11, find 'r'
5.
1. Prove that
n+1
2. Show that
10
Pr+1 = (n + 1) nPr
P3 = 9P3 + 3 9P2
iv)
15
P1
v)
38
P0
11 990
10 90
9
9
1
Permutations and Combinations
79
Practical Problems on Permutation
Example 1. 6 songs are to be rendered in a programme. In how many different orders
could they be performed?
Sol. The different orders in which 6 songs that can be performed is 6P6 = 6!
6! = 6 × 5 × 4 × 3 × 2 × 1 [ FPC]
= 720 different orders
n!
n r !
6!
6
6 6 ! =
3 2 1
= 720
1
6 songs can be rendered in a programme in 720 different orders.
or
n
Pr
6
P6
5
4
Example2. How many words (with or without dictionary meaning) can be made from
the letters in the word LASER assuming that no letter is repeated it, such that
(i) All letters are used at a time
(ii) 3 letters are used at a time
(iii) All letters are used such that it should begin with letter A and end with letter R
Sol. Number of letters in the word LASER is 5.
(i) If all letters are used at a time, then n = 5, r = 5
The number of words that can be formed
5
= P5
5
5!
5 !
5!
0!
5!
1
5 ! 120
120 words can be formed with or without meaning.
(ii) If 3 letters are used at a time, then n = 5, r = 3
 The number of words formed with or without meaning
5
= P3
5!
5 3 !
5
4
3
2
2
1
1
60
So there will be 60 words with or without meaning.
(iii)
Box 1
Box 2 Box 3
letter A is fixed
Box 4
Box 5
letter R is fixed
There are 5 letters in the given word LASER of which two letters A and R are
fixed in the first & last place respectively.
So only three boxes have to be filled namely 2nd, 3rd and 4th. This can be done in
3
P3
3
3!
3 !
3
2 1
0!
6
1
6 ways
 There will be only 6 words with or without meaning which begins with letter
'A' and end with the letter 'R'.
80
UNIT-4
Example 3. In how many ways can 7 different books be arranged on a shelf? In how
many ways three particular books are always together?
Sol. The 7 books can be arranged in
7
P7 = 7!
=7×6×4×3×2×1
= 5040 ways.
Since three particular books are always together, let us tie three books together
and then consider them as one book (or one unit). Remaining four books have to be
considered separately. So in all we can consider 7 books as
[4 books +
1 book
] = 5 books
(1 unit of 3 books)
These 5 books can be arranged in 5P5 = 5! = 5 × 4 × 3 × 2 × 1 = 120 ways.
In each of these 120 ways, the three particular books can be arranged in
3! = 3 × 2 × 1 = 6 ways { (B1 B2 B3)(B1 B3 B2), (B2 B1 B3), (B2 B3 B1), (B3 B1 B2)(B3 B2 B1)
By FPC, the total number of arrangements 5P5 × 3! = 120 × 6 = 720 ways.
So, the number of ways of arranging 7 books, so that three particular books is never
together is 5040 – 720 = 4320
Example4.
Sol.
How many five digit numbers can be formed with the digits {2, 3, 5, 7, 9}
which lie between 30,000 and 90,000 using each digit only once.
Ten thousand
place
3 ways {3,5 or 7}
Thousand
place
4 ways
Hundred
place
3 ways
Tenth
place
2 ways
unit 's
place
1way
Since we require numbers which are greater than 30,000 and less than 90,000,
the ten thousandth place can be filled by {3, 5 or 7} only.
Therefore ten thousandth place can be filled in by 3 ways only. The remaining four
places can be filled with the remaining four digits. This can be done is 4! ways.
Therefore the total number of numbers lying between
30,000 and 90,000 is 3 × 4! { FPC}
= 3 × 4 × 3 × 2 × 1 = 72
Example 5. A DNA molecule will have a nitrogen base which
consists of different bases A, G, T or C all attached to it? In
how many ways the three bases can be arranged without
repetition.
Sol. Here, n = 4 and r = 3. Hence the number of ways
4
P3
4!
4 3 !
4
3
2 1
1
24 ways
A = Adenine
C = Cytosine
G = Genanine
T = Thymine
Permutations and Combinations
81
EXERCISE 4.5
1. How many words with or without dictionary meaning can be formed using all the
letters of the word 'JOULE' using each letter exactly once?
2. It is required to seat 5 men and 4 women in a row so that the women occupy the
even places. How many such arrangements are possible?
3. In how many ways can 6 women draw water from 6 wells, if no well remains unused?
4. Seven students are contesting the election for the Presidentship of the student's
union. In how many ways can their names be listed on the ballot papers?
5. 8 students are participating in a competition. In how many ways can the first three
prizes be won?
6. Find the total number of 2- digit numbers.
7. There are 5 stickers of different sizes. It is desired to make a design by arranging
them in a row. How many such designs are possible?
8. How many 4- digit numbers can be formed using the digits 1, 2, 3, 7, 8, 9 (repetitions
not allowed)?
(a) How many of these are less than 6000?
(b) How many of these are even?
(c) How many of these end with 7?
9. There are 15 buses running between two towns. In how many ways can a man go to
one town and return by a different bus?
To find out the number of combinations of 'n' distinct objects taken 'r' at a time where
0rn
Consider the following examples of selecting and arranging letters from a, b, c and d.
Here n = 4:
Combinations
Permutations
r = 1,
a, b, c, d
a, b, c, d
4
4
C1 = 4
P1 = 4
4
Each combination of C1 give rise to 1! Permutation.
r=2
ab,ac,ad,ba,bc,bd
ab,ac,ad,bc,bd,cd
ca,cb,cd,da,db,dc
4
C2 = 6
4P2 = 12
Each combination of 4C2 give rise to 2! Permutations.
r = 3 abc,abd,acd,bcd
abc abd bcd acd
4C3 = 6
acb adb bdc adc
bac bad cbd cad
bca bda cdb cda
cab dab dbc dac
cba dba dcb dca
4P3 = 24 ways
Each combination of 4C3 give rise to 3! Permutations.
82
UNIT-4
r = 4 abcd
4
C4 = 1
abcd bacd cabd dabc
abdc badc cadb dacb
acbd bcad cdba dbac
acdb bcda cdab dbca
adcb bdac cdab dcab
abdc bdca cbda dcba
 4P4: 24
Each combination in 4C4 gives rise to 4! Permutations.
In General:
Each combination in nCr give rise to r! permutations.
How many permutations can there be altogether?
n
 Total no of Permutations = r ! × nCr
i.e. nCr = n

n
Pr = r ! × nCr
 nCr =
Pr
r!
n!
r !r !
Therefore, the number of combinations of 'n' distinct objects, taken 'r' at a time is
given by
n
Cr
r!
n!
n-r !
n
Note: The relationship between nPr and nCr is given by nCr =
Study the various expansions of nCr
n
1.
2.
n
Cr
n
n
3.
4.
n
Cr
Cr
Cr
Pr
r!
n!
r! n r !
n n
1 n
2
n n
1 n
2
............... 3
r! n r !
................
2 1
n
r
1
r!
Now let us consider some special cases of nCr
Case (i): Let r = 0
n
If r = 0 then, C0

100
n!
0! n 0
n!
1 n!
C0 = 1, 500C0 = 1, 1000C0 = 1
1  nC = 1
0
Pr
r!
Permutations and Combinations
83
Case (ii): Let r = 1
n
If r = 1 then, C1
n n 1 !
n!
=
1! n 1 ! 1 n 1 !
nC1 = n
Example:
100
200
C1 = 100,
357
C1 = 200
C1 = 357
Case (iii): Let r = n
n
If r = n then, Cn
n!
n! n n !
n!
n! 1
n!
n !0!
1
nCn = 1
Example:
100
C100 = 1,
789
1497
C789 = 1,
C1497 = 1
Remember :
n
n!
n r !r !
Cr
If r = 0, nC0 = 1, If r = 1, nC1 = n, If r = n, nCn = 1
ILLUSTRATIVE EXAMPLES
Evaluate : (i) 6C4
Example 1.
Sol.
n
(i)
6
Cr
C4
(ii) 8C3
n!
n r !r !
6!
6 4 !4!
6!
2!4!
8!
8 3 !3!
8!
5!3!
3
4
(ii)
8
C3
6
5
2
1
8
7
5!
4!
4!
2
6
15
5!
31 2
56
1
Example2.If nC2 = 10, find n
Sol. nC2 = 10
n
Cr
n
n!
r !r !
n
 C2
n
n n 1
10
2
 n(n – 1) = 20 5(5 – 1) = 20
n!
2 !2!

1 n
n n
n
2 !
2 ! 2!
n
 C2
 5 × 4 = 20
 n=5
84
UNIT-4
Example 3. If 6Pr = 360 and 6Cr = 15, find 'r'.
6
Sol.
Pr = 6Cr × r !
360 = 15 × r !
360
15
 r!
24 = 1 × 2 × 3 × 4 = 4!
r=4
Example 4. Prove that nCr = nCn – r
n!
r! n r !
n
Sol. We know that Cr
.........(i)
Replacing r by (n – r) in the equation (i) we get
n
n
n
Cn
r
Cn
r
Cn
r
n!
r ! n n
n
r
!
n!
n
r ! n
r !
n
n!
n r !r !
........(ii)
By comparing (i) and (ii) we can say that
n
Cr = nCn – r
Example 5. If nC8 = nC12, find n.
Sol. nC8 = nC12
n
C8 = nCn – 12
 nCr
n
Cn
r
8 = n –12 n = 12 + 8 = 20.
EXERCISE 4.6
1. Evaluate (i)
n
10
C3
(ii)
60
C60
n
2. (i) If C4 = C7 find n.
(iii)
n
4. Verify that 8C4 + 8C5 = 9C4
n
n 1
Cr
Cr
1
n
when 1
r
C97
(ii) If Pr = 840, nCr = 35, find n.
3. If 2nC3 : nC3 = 11:1, find n.
5. Prove that
100
r
6. If nCr-1: nCr: nCr +1 = 3 : 4 : 5, find r.
n.
Permutations and Combinations
85
Example 1. A man has 6 friends. In how many ways can he invite one or more of
them to a party?
Sol. The different ways of combinations of inviting 6 friends are inviting exactly one,,
exactly 2, ........... exactly 6. These can be done in
6
C1 , 6C 2 , 6C3 , 6C 4 , 6C 5 , 6C 6 ways.
The total number of ways in which one or more can be invited
= 6 C1
6
C2
6
= 6 C1
6
C2
6
C3
C3
6
C4
6
C5
6
6
C2
6
C1
6
C6
C0
= 6 + 15 + 20 + 15 + 6 + 1 = 63 ways.
Example 2. For a set of 5 true or false questions no student has written all correct
answers and no two students have given the same sequence of answers. What is
the maximum number of students in the class for this to be possible?
Sol.
Each question can be answered in exactly 2 ways either T or F.
 Total number of ways in which the 5 questions can be answered
= 2 × 2 × 2 × 2 × 2 = 32
Out of these 32 ways exactly one will be all correct answer.
Since no student has written all correct answers
The maximum possible number of students = 32 – 1 = 31.
Example 3. 4 friends shake hands mutually. Find the number of handshakes.
Sol. Let the friends be A, B, C and D. For one hand shake, there will
be 2 persons involved. To start with let person 'A' shakes hand
with the person 'B'. It is the same as person 'B' shaking hands A
with person 'A'. AB and BA are the same.
So in finding the number of handshakes, we observe that order {AB, AC, AD}
is not important.
Hence the number of handshakes will be always
B
n n 1
n
n
C2, and C2 =
.
2
{BC, BD}
In this problem n = 4.  Number of handshakes will be 4C2.
4
C2
4!
4 2 !2!
2
4 3 2 1
2 1 2 1
6
C
{CD}
B
C
D
C
D
D
Example 4. Everybody in a function shakes hand with everybody else. The total
number of handshakes is 45. Find the number of persons in the function.
Sol.
Let the number of persons in the function be n.
Then nC2 = 45
n
n!
2 !2!
n n 1
2
45
45
86
UNIT-4
n(n – 1) = 90
n(n – 1) = 10 × 9
n(n – 1) = 10(10 – 1)
 n = 10
Hence the number of persons in the function is 10.
Example 5. How many committees of five with a given chairperson can be selected
from 12 persons?
Sol.
The chairperson can be choosen in 12 ways and following this the other four on
the committee can be choosen in 11C4 ways.
 The possible number of such committees = 12 ×
11
C4
= 12 × 330 = 3960.
Example 6. There are 8 points such that any of 3 of them are non collinear. How
many straight lines can be drawn by joining these points?
Sol. A straight line is obtained by joining 2 points. Let the points be A and B.
The straight line is obtained by joining either A with B or B with A.
 AB and BA are same.
 This is a problem on combination.
A
B
 Total number of lines that can be drawn out of 'n' non-collinear points = nC2
Verification :
1
n
1
C
n
n
n!
We know,
n
2
Cr
2
n r !r !
Here, n = 8, r = 2
8!
8
C2
1
8
2 !2!
8 8 1 = 28
 8C2 =
4
2
8 7 6!
 28 straight lines can be drawn by joining 8 points.
6! 2
8
 C 2 28
Example 7. There are 10 points such that any 3 of them are noncollinear. How many
triangles can be formed by joining these points?
Sol. A triangle is formed by joining 3 non-collinear points.
Here the order in which the points are joined is not at all important.
 Total number of triangles that can be drawn out of 'n' non-collinear points = nC3
Here, n = 10, r = 3,
10
C3
10!
10 3 !3!
n
Cr
n
n!
r !r !
3
4
10!
10 9
8
7! 3! 
7! 3 2
7!
1
= 120
 120 triangles can be formed by joining 3 points out of 10 non-collinear points.
Permutations and Combinations
87
Alternate method:
n
C3
n
 C3
n!
n r !r !
n n
n!
n 3 !3!
1 n
6
n n
n
1 n
3 !
2 n
3
3 !
2 1
2
A
If n = 10
10
C3
10 10 1 10
2
10
3
9
63
6
8
4
= 120
B
C
Example 8. How many diagonals can be drawn in a hexagon?
Sol. A hexagon has 6 vertices  n = 6
A diagonal is obtained by joining the opposite vertices in pairs.
 Total number of sides and diagonals = 6C2
6 6 1
n n 1
6
 C2
2
2
15 lines includes 6 sides.
n
C2
= 15
 Number of diagonals = 15 – 6 = 9
Alternate method:
Number of diagonals = Number of straight lines formed – Number of sides of polygon
1
n n 1 n
= nC2 – n
=
2
n2
=
n 2n
2
n2
n n 3
2
3n
2
 Number of diagonals in a polygon of n sides =
In an hexagon, n = 6
 Number of diagonals
n n 3
2
n n 3
2
6 6 3
2
= 9
Example 9. The maximum number of diagonals in a polygon is 14. Find the number
of sides.
Sol.
Number of diagonals in a polygon
14
n n 3
2
n n 3
2
 n(n – 3) = 14 × 2
 n(n – 3) = 7(7 – 4)
n=7
 n(n – 3) = 7 × 4
88
UNIT-4
n n 3
2
Alternate : 14
 n2 – 3n – 28 = 0
n(n – 3) = 14 × 2
 n2 – 7n + 4n – 28 = 0
n(n – 7) + 4(n – 7) = 0  (n – 7) (n + 4) = 0
n–7=0
OR
n+4=0
n=7
n=–4
'n' cannot be a negative number
Hence n = 7 (Heptagon)
Remember
When 'n' number of non-collinear points are given in a plane,
n n 1
2
n
* number of straight lines = C 2
n
* number of diagonals in a polygon = C2
1 n
6
n n
n
* number of triangles = C 3
2
n n 3
2
n
Pascal Triangle
Pascal, Blaise (1623 - 1662) great French Mathematician , probabilist, combinatiorist,
physicist and philosopher.
Observe the triangle pattern discovered by Pascal.
Compare the corresponding elements in the two triangles.
The triangle on the right side is called Pascal triangle.
1
C0
1
1
C0
2
4
5
5
C0
6
C0
6
C1
3
4
C1
6
C2
4
C2
5
C1
3
C2
5
C2
6
C3
C3
C4
5
C4
1
5
C4
6
2
1
4
C3
C3
1
1
C2
C1
4
C0
1
2
C1
3
C0
6
C1
2
C0
3
1
6
C5
1
C5
6
C
1
3
4
3
15
1
6
10
5
6
1
4
10
20
1
1
5
15
6
1
Permutations and Combinations
89
EXERCISE 4.7
1. Out of 7 Consonants and 4 vowels, how many words of 3 consonants and 2 vowels
can be formed?
2. In how many ways can 5 sportsmen be selected from a group of 10?
3. In how many ways a cricket team of eleven be selected from 17 players in which
5 players are bowlers and the cricket team must include 2 bowlers?
4. How many (i) lines (ii) triangles can be drawn through 8 points on a circle?
5. How many diagonals can be drawn in a (i) decagon (ii) icosagon
6. A Polygon has 44 diagonals. Find the number of sides.
7. There are 3 white and 4 red roses in a garden. In how many ways can 4 flowers of
which 2 red be plucked?
8. In how many ways can a student choose 5 courses out of 9 courses, if 2 courses are
compulsory for every student?
9. There are 5 questions in a question paper. In how many ways can a boy solve one or
more questions?
10. In how many ways 4 cards from a pack of 52 playing cards can be chosen?
11. In an election there are 7 candidates and three are to be elected. A voter is allowed
to vote for any number of candidates not greater than the number to be elected. In
how many ways can we vote?
EXERCISE 4.8
1. If there are 6 periods in each working day of a school, in how many ways can one
arrange 5 subjects such that each subject is allowed at least one period?
2. A committee of 5 is to be formed out of 6 men and 4 ladies. In how many ways can
this be done when
(i) at least 2 ladies are included. (ii) at most 2 ladies are included.
3. A sports team of 11 students is to be constituted choosing at least 5 from class IX
and at least 5 from class X. If there are 8 students in each of these classes, in how
many ways can the team be constituted?
4. In a vegetable mela or fair, an artist wants to make mascot (a toy that represents
an organization) with Beans, Carrot, Peas and Tomato in a line.
In how many ways can the artist make the mascot?
5. From a group of 12 students, 8 are to be chosen for an excursion. There are 3
students who decide that either of them will join or none of them will join. In how
many ways can the 8 be chosen?
6. How many chords can be drawn through 20 points on a circle?
7. The English alphabet has 5 vowels and 21 consonants. How many words with 2
different vowels and 2 different consonants can be formed from the alphabet?
8. In how many ways 5 letters be posted in 3 post boxes, if any number of letters can be
posted in all of the three post boxes?
9. In a group of 15 students there are 6 scouts. In how many ways can 12 students be
selected, so as to include at least 4 scouts?
90
UNIT-4
Permutation and Combination
Permutation
n
Pr
Fundamental
Principle of
Counting (FPC)
Combination
n
Cr
n!
n
Factorial
notation n!
Pr =
(n r )!
where 0 r  n
n
If r = 0,
n
P0 = 1
n
If r = n,
n
Pn = n!
If r = 1,
n
P1 = n
Cr =
n!
r !(n r )!
Pr = nCr×r!
If r = 0,
n
C0 = 1
If r = n,
n
Cn = n
If r = 1,
n
C1 = n
ANSWERS
EXERCISE 4.1
1] 120
EXERCISE 4.3
1] (i) 7!
2] 12
(ii) 18!
3] 720
4] 336
9!
(iv) 2 × 4!
5!
(vi) 870
3] (i) 210
6] n(n + 1)
(iii)
2] (i) 720
5] 8
6] 20
(ii) 3,62,880 (iii) 40,200
(iv) 42
(v) 220
(ii) 105
4] 720
5] (i) 3
(ii) 121
EXERCISE 4.4
1] (i) 11,880
(ii) 5,550
(iii) 40,320
(iv) 15
(v) 1
2] (i) 7
(ii) 15
(iii) 3
3] (i) 6
(ii) 8
4] (i) 5
(ii) 5
EXERCISE 4.5
1] 120
2] 2,880
3] 720
4] 1
5] 336
6] 90
7] 120
8] (a) 180
(b) 120
(c) 60
9] 210
EXERCISE 4.6
1] (i) 120
(ii) 1
(iii) 1,61,700
2] (i) 11
(ii) 7
3] 6
6] 27
EXERCISE 4.7
1] 25,200
2] 252
3] 2,200
4] (i) 28
(ii) 56
5] (i) 35
(ii) 170
6] 11
7] 18
8] 35
9] 31
10] 2,70,725
11] 63
EXERCISE 4.8
1] 3,600
2] (i) 186
(ii) 186
3] 3,136
4] 24
5] 117
6] 190
7] 50,400
8] 243
9] 435