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7.1 Introduction 9!
Example 3: Evaluate
6!3!
The factorial notation was introduced by Christian Kramp (1760 - 1826) in 1808. This
notation will be frequently used in this chapter as well as in inding the Binomial Coeicients
= = 84
9! (9.8.7)6!
in later chapter. Let us have an introduction of factorial notation. Solution:
Let n be a positive integer. Then the product n(n - 1)(n - 2). . . . 3 . 2 . 1 is denoted by
6!3! 6!(3.2.1)

n! or In and read as n factorial.

=
=
9! 9.8.7.6.5.4.3.2.1
or 84

n! = n(n - 1)(n - 2) ....3.2.1

6!3! 6.5.4.3.2.1.3.2.1
That is,

For Example, Exercise 7.1

1! = 1
2! = 2.1 =2 ⇒2! = 2.1! 1. Evaluate each of the following:
3! = 3.2.1 =6 ⇒3! = 3.2!
4! = 4.3.2.1 = 24 ⇒ 4! = 4.3!
⇒5! = 5.4!
8! 10!
5! = 5.4.3.2.1 = 120 i) 4! ii) 6! iii) iv)
⇒6! = 6.5!
7! 7!
and 6 ! = 6.5.4.3.2.1 = 720
11! 6! 8! 11!
v) vi) vii) viii)
Thus for a positive integer n, we deine n factorial as: 4!7! 3!3! 4!2! 2!4!5!
n! =n(n - 1)! where 0!= 1

2!( 9 - 2 )! 15!(15 - 15 )!
9! 15! 3!
8! ix) x) xi) xii) 4!.0!.1!
Example 1: Evaluate 0!
6!

==
8! 8.7.6.5.4.3.2.1 2. Write each of the following in the factorial form:
Solution: 56
6! 6.5.4.3.2.1 i) 6.5.4. ii) 12.11.10 iii) 20.19.18.17

10.9 8.7.6 52.51.50.49

iv) v) vi)
Example 2: Write 8.7.6.5 in the factorial form 2.1 3.2.1 4.3.2.1

(n + 1)(n)(n - 1)
8.7.6.5.4.3.2.1 8! vii) n(n - 1 )(n - 2) viii) (n + 2)(n + 1)(n) ix)
Solution: 8.7.6.5 = = 3.2.1
4.3.2.1 4!
x) n(n - 1)(n - 2)....(n - r + 1)

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are two ways of writing at second place as shown. 3 2

7.2 Permutation
Now just one vertex is left. So, we can write at third place only one vertex in one way
as shown. 3 2 1
Suppose we like to ind the number of diferent ways to name the
The total number of possible ways (arrangements) is the product 3.2.1= 6.
triangle with vertices A, B and C.
This example illustrates the fundamental principle of counting.
The various possible ways are obtained by constructing a tree
diagram as follows:
Fundamental Principle of Counting:
Suppose A and B are two events. The irst event A can occur in
p diferent ways. After A has occurred, B can occur in q diferent ways.
The number of ways that the two events can occur is the product p.q.

This principle can be extended to three or more events. For instance, the number of
ways that three events A, B and C can occur is the product p.q.r.
One important application of the Fundamental Principle of Counting is to determine
the number of ways that n objects can be arranged in order. An ordering (arrangement) of
n objects is called a permutation of the objects.
A permutation of n diferent objects is an ordering (arrangement) of the objects
such that one object is irst, one is second, one is third and so on.
To determine the possible ways, we count the paths of the tree, beginning from the
According to Fundamental Principle of Counting:
start to the end of each branch. So, we get 6 diferent names of triangle.
i) Three books can be arranged in a row taken all at a time = 3.2.1 = 3! ways
ABC, ACB, BCA, BAC, CAB, CBA.
ii) Number of ways of writing the letters of the WORD taken all at a time
Thus there are six possible ways to write the name of the triangle with vertices A, B and
= 4.3.2.1 = 4!
C.
Each arrangement is called a permutation. Now we have the following deinition.
Explanation: In the igure, we can write any one of the three vertices A, B, C at irst place.
A permutation of n diferent objects taken r (7 n) at a time is an arrangement of
After writing at irst place any one of the three vertices, two vertices are left. So, there are
the r objects. Generally it is denoted by n Pr or P(n,r).
two choices to write at second place. After writing the vertices at two places, there is just one

Prove that: n pr= n(n - 1)(n - 2)....(n - r + 1)=

( n - r )!
vertex left. So, we can write only one vertex at third place. n!

Another Way of Explanation:

Proof: As there are n diferent objects to ill up r places. So, the irst place can be illed in n ways.
Think of the three places as shown
Since repetitions are not allowed, the second place can be illed in (n-1) ways, the third place
is illed in (n-2) ways and so on. The rth place has n - (r - 1) = n - r + 1 choices to be illed in.
Since we can write any one of the three vertices at irst place, so it is written in 3 diferent
ways as shown. 3
Therefore, by the fundamental principle of counting, r places can be illed by n diferent objects
in n(n - 1)(n - 2) .... (n - r + 1) ways
Now two vertices are left. So, corresponding to each way of writing at irst place, there
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∴ p = n(n - 1)(n - 2)....(n - r + 1)=

Example 3: In how many ways can a set of 4 diferent mathematics books and 5 diferent
( ) physics books be placed on a shelf with a space for 9 books, if all books on the same subject
are kept together?
n(n - 1)(n - 2)....(n - r + 1)(n - r)(n - r - 1) .... 3.2.1
=
(n - r )(n - r - 1) .... 3.2.1 Solution: 4 diferent Mathematics books can be arranged among themselves in 4! ways. 5
⇒ n Pr =
( n - r )!
n! diferent Physics books can be arranged among themselves in 5! ways.To every one way of
arranging 4 mathematics books there are 5! ways of arranging 5 physics books. The books
which completes the proof. in the two subjects can be arranged subject-wise in 2! ways.

Corollary: If r = n, then
So the number of ways of arranging the books as given by.
p=n = = = n!
( n - n )! 0! 1
n n! n! n!
4! % 5! % 2! = 4 % 3 % 2 % 1 % 5 % 4 % 3 % 2 % 1 % 2 % 1
= 5740

⇒ n diferent objects can be arranged taken all at a time in n! ways. Exercise 7.2

Example 1: How many diferent 4-digit numbers can be formed out of the digits 1, 2, 3, 4, 1. Evaluate the following:
5, 6, when no digit is repeated?
i) 20 P3 ii) 16 P4 iii) 12
P5 iv) 10
P7 v) 9
P8
2. Find the value of n when:
P2 = 30 Pn = 11.10.9 P4 : n-1P3 = 9 :1
Solution: The total number of digits = 6 n 11 n
i) ii) iii)
The digits forming each number = 4.
So, the required number of 4-digit numbers is given by:
3. Prove from the irst principle that:

p= = = = 6.5.4.3
= 360 i) n Pr = n . n-1Pr -1 ii) = n -1
Pr + r. n-1Pr -1
( 6 - 4 )! 2!
n
6 6! 6! 6.5.4.3.2.1 Pr
4
2.1 4. How many signals can be given by 5 lags of diferent colours, using 3 lags at a time?
Example 2: How many signals can be made with 4-diferent lags when any number of them 5. How many signals can be given by 6 lags of diferent colours when any number of
are to be used at a time? lags can be used at a time?
6. How many words can be formed from the letters of the following words using all letters
Solution: The number of flags = 4 when no letter is to be repeated:
i) PLANE ii) OBJECT iii) FASTING?
Number of signals using 1 lag = 4 P1 = 4
7. How many 3-digit numbers can be formed by using each one of the digits 2, 3, 5, 7, 9
Number of signals using 2 lags = 4 P2 = 4 . 3 =12
only once?
Number of signals using 3 lags = 4 P3 = 4.3.2 = 24 8. Find the numbers greater than 23000 that can be formed from the digits 1, 2, 3, 5, 6 ,

∴
Number of signals using 4 lags = P4 = 4.3.2.1 = 24
4
without repeating any digit.
Total Number of signals = 4 + 12 + 24 + 24 = 64. HINT: The irst two digits on L.H.S. will be 23 etc.

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9. Find the number of 5-digit numbers that can be formed from the digits 1, 2, 4, 6, 8 alike of second kind and the rest of them are all diferent. Let x be the required number of
(when no digit is repeated), but permutation. Replacing n1 alike things by n1diferent things and n2 alike things by n2 diferent
i) the digits 2 and 8 are next to each other; things, we shall get all the n things distinct from each other which can be permuted among
ii) the digits 2 and 8 are not next to each other. themselves in n! ways. As n1 diferent things can be permuted among themselves in (n1)!
10. How many 6-digit numbers can be formed, without repeating any digit from the digits ways and n2 diferent things can be arranged among themselves in (n2)! ways, so because of
0, 1, 2, 3, 4, 5? In how many of them will 0 be at the tens place? the replacement suggested above, x permutation would increase to x x (n1) ! % (n2)! number
11. How many 5-digit multiples of 5 can be formed from the digits 2, 3, 5, 7,9, when no digit of ways.
is repeated. ∴ x × (n1 )!× (n 2 )! =
(n)!
In how many ways can 8 books including 2 on English be arranged on a shelf in such a
 n 
12.
way that the English books are never together? = =  
(n)!
(n1 )!× (n2 )!  n1 , n2 
Hence x
13. Find the number of arrangements of 3 books on English and 5 books on Urdu for
placing them on a shelf such that the books on the same subject are together.
Cor. If there are n1 alike things of one kind, n2 alike things of second kind and n3 alike
14. In how many ways can 5 boys and 4 girls be seated on a bench so that the girls and the
things of third kind, then the number of permutation of n things,taken all at a time is given
boys occupy alternate seats?
by:

 n 
= 
7.2.1 Permutation of Things Not All Different

n!
Suppose we have to ind the permutation of the letters of the word BITTER, using all
(n1 )!× (n2 )!× (n3 )!  n1 , n2 n3 
the letters in it. We see that all the letters of the word BITTER are not diferent and it has 2
Ts in it. Obviously, the interchanging of Ts in any permutation, say BITTER, will not form Example 1: In how many ways can be letters of the word MISSISSIPPI be arranged when
a new permutation. However, if the two Ts are replaced by T1 and T2, we get the following all the letters are to be used?
two permutation of BITTER
BIT1T2ER and BIT2T1ER Solution: Number of letters in MISSISSIPPI = 11
In MISSISSIPPI,
Similarly, the replacement of the two Ts by T1 and T2 I is repeated 4 times
in any other permutation will give rise to 2 permutation. S is repeated 4 times
P is repeated 2 times
Now, BIT1T2ER consists of 6 diferent letters which can be permuted among themselves M comes only once.
in 6 ! diferent ways. Hence the number of permutation of the letters of the word BITTER
 11 
Required number of permutation =  
taken all at a time
= = =
6! 6.5.4.3.2.1
360  4,4,2,1
2 2

=
=
This example guides us to discover the method of inding the permutation of n things (11)!
(4)!× (4)!× (2)!× (1)!
34650 ways
all of which are not diferent. Suppose that out of n things, n1 are alike of one kind and n2 are
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7.2.2 Circular Permutation

So far we have been studying permutation of things which can be represented by

the points on a straight line. We shall now study the permutation of things which can be
represented by the points on a circle. The permutation of things which can be represented
by the points on a circle are called Circular Permutation.
Figure (i) Figure (ii)
The method of inding circular permutation is illustrated by the following examples.
By lipping the necklace we get the necklace as shown in igure (ii). We observe that the
Example 2: In how many ways can 5 persons be seated at a round table. two arrangements of the beads are actually the same.

Hence the required number of necklaces = = × (7)! =

1
2520

Solution: Let A, B, C, D, E be the 5 persons One of the ways of seating

2
Exercise 7.3
them round a table is shown in the adjoining igure. If each person
moves one or two or more places to the left or the right, they will, no 1. How many arrangements of the letters of the following words, taken all
doubt, be occupying diferent seats, but their positions relative to each together, can be made:
other will remain the same. i) PAKPATTAN ii) PAKISTAN
ii) MATHEMATICS iv) ASSASSINATION?
So, when A occupies a certain seat, the remaining 4 persons will be permuting their seats 2. How many permutation of the letters of the word PANAMA can be made, if P is to be
among themselves in 4! ways. the irst letter in each arrangement?
Hence the number of arrangements = 4! = 24 3. How many arrangements of the letters of the word ATTACKED can be made, if each
arrangement begins with C and ends with K?
Example 3: In how many ways can a necklace of 8 beads of diferent colours be made? 4. How many numbers greater than 1000,000 can be formed from the digits 0, 2, 2,2, 3,
4 ,4 ?
Solution: The number of beads = 8 5. How many 6-digit numbers can be formed from the digits 2, 2, 3, 3, 4, 4? How many
The number of arrangements of 8 beads in the necklace will be like the seating of 8 of them will lie between 400,000 and 430,000?
persons round a table.
⇒ The number of such necklaces (ixing one of the beads) = 7!
6. 11 members of a club form 4 committees of 3, 4, 2, 2 members so that no member is
a member of more than one committee. Find the number of committees.
Now suppose the beads are a, b, c, d, e , f g , h and the necklace is as shown in Fig. (i) 7. The D.C.Os of 11 districts meet to discuss the law and order situation in their
below: districts. In how many ways can they be seated at a round table, when two particular
D.C.Os insist on sitting together?
8. The Governor of the Punjab calls a meeting of 12 oicers. In how many ways can they
be seated at a round table?
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9. Fatima invites 14 people to a dinner. There are 9 males and 5 females who are seated

⇒ cr ×
= ∴=
( n - r )! r !( n - r )!
at two diferent tables so that guests of one sex sit at one round table and the guests n n! n n!
r cr
of the other sex at the second table. Find the number of ways in which all gests are
seated.
Which completes the proof.
10. Find the number of ways in which 5 men and 5 women can be seated at a round table
Corollary:
in such a way that no two persons of the same sex sit together.
In how many ways can 4 keys be arranged on a circular key ring?
= = =
n!( n - r )! n!0!
11. n! n!
n
12. How many necklaces can be made from 6 beads of diferent colours? i) If r = n, then cn 1

= = = 1
7.3 Combination
0!( n - 0 )!
n n! n!
ii) If r = 0, then c0
0!n!
While counting the number of possible permutation of a set of objects, the order is
important. But there are situations where order is immaterial. For example
ABC, ACB, BAC, BCA, CAB, CBA are the six names of the triangle whose vertices
i)
7.3.1 Complementary Combination
are A, B and C. We notice that inspite of the diferent arrangements of the vertices
Prove that: nCr = nCn-r
Proof: If from n diferent objects, we select r objects then (n - r) objects are left.
of the triangle, they represent one and the same triangle.
ii) The 11 players of a cricket team can be arranged in 11! ways, but they are players
Corresponding to every combination of r objects, there is a combination of (n - r)
of the same single team. So, we are interested in the membership of the committee
objects and vice versa.
(group) and not in the way the members are listed (arranged). Therefore, a
Thus the number of combinations of n objects taken r at a time is equal to the number
combination of n diferent objects taken r at a time is a set of r objects.
of combinations of n objects taken (n - r) at a time.
The number of combinations of n diferent objects taken r at a time is denoted by nCr
∴ n
Cr =
n
Cn - r
n
or C(n, r) or r  and is given by
 
cn-r =
(n - r )!( n - n + r )!
n n!
Other wise:

cr =
r !( n - r )! = =
n!

( n - r )!r! r !( n - r )!
n
n! n!

Proof: There are nCr combinations of n diferent objects taken r at a time. Each combination ⇒ ncn-r =
n
cr
consists of r diferent objects which can be permuted among themselves in r! ways. So, each
combination will give rise to r! permutation. Thus there will be nCr x r! permutation of n cr r > .
n
Note: This result will be found useful in evaluating nCr when
diferent objects taken r at a time. 2
n
cr × r !=n pr
c2 =
(12).(11)
e.g l2
C10 = 12
C12-10 = 12
= (6). (11) =66
2
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Example 1: If nC8 = nC12, ind n. = R.H.S.

Hence n-1Cr + n-1Cr- 1 = nCr

∴
n
Solution: We know that Cr = nCn - r
n
C8 = nCn-8 (i) Exercise 7.4
But it is given that nC8 = nC12 (ii)
From (i) and (ii), we conclude that 1. Evaluate the following:
12
n
Cn-8 = n
C12 i) C3 ii) 20C17 iii) n
C4
⇒ n - 8 = 12
∴
2. Find the value of n, when
n = 20
12 × 11
i) n
C5 = nC4 ii) n
C10 = iii) C12 = nC6
n
2!
Example 2: Find the number of the diagonals of a 6-sided igure.
3. Find the values of n and r, when
Solution: A 6-sided igure has 6 vertices. Joining any two vertices we get a line segment. i) nCr = 35 and nPr = 210
ii) n-1Cr-1 : nCr : n+1Cr+1 = 3 : 6 : 1 1
∴ Number of line segments = = 15
6 6! 4. How many (a) diagonals and (b) triangles can be formed by joining the vertices of the
C2
2!4! polygon having:
But these line segments include 6 sides of the igure i) 5 sides ii) 8 sides iii) 12 sides?
∴ Number of diagonals = 1 5 - 6 = 9 5. The members of a club are 12 boys and 8 girls. In how many ways can a committee of 3
boys and 2 girls be formed?
Example 3: Prove that: n-1
Cr + n-1Cr- 1 = nCr 6. How many committees of 5 members can be chosen from a group of 8 persons when
each committee must include 2 particular persons?
Solution: L.H.S. = n-1
Cr + n-1Cr- 1 7. In how many ways can a hockey team of 11 players be selected out of 15 players? How
many of them will include a particular player?
n -1 n -1
= +
r n -1- r r -1 n - r
8. Show that: l6C11 + 16C10 = 7C11
9. There are 8 men and 10 women members of a club. How many committees of can be

n -1 n -1
formed, having;
= +
r r -1 n - r -1 r - 1(n - r ) n - r - 1
i) 4 women ii) at the most 4 women iii) atleast 4 women?
10. Prove that nCr + nCr-1 = n+1Cr.

n -1 1 1  n -1 n - r + r 
= = + =
 
r -1 n - r -1  r n - r  r - 1 n - r - 1 r (n - r ) 
7.4 Probability

n n -1
= = =
n We live in an uncertain world where very many events cannot be predicted with
r r - 1(n - r ) n - r - 1 r n - r
n
Cr
complete certainty, e.g.
i) In a cloudy weather, we cannot be sure whether it will or will not rain. However,
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