Permutations & Combinations JEE Notes | EduRev

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JEE : Permutations & Combinations JEE Notes | EduRev

 Page 1


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N O D E 6 \ E \ D a t a \ 2 0 1 4 \ K o t a \ J E E - A d v a n c e d \ S M P \ M a t h s \ U n i t # 1 2 \ E N G \ 0 1 . P E R M U T A T I O N - C O M B I N A T I O N . p 6 5
J E E - M a t h e m a t i c s
1 . FUNDAMENTAL PRINCIPLE OF COUNTING  (counting without actual counting):
If an event A can occur in 'm' different ways and another event B can occur in 'n' different ways, then the total
number of different ways of-
(a) simultaneous occurrence of both events in a definite order is m× n. This can be extended to any number
of events (known as multiplication principle).
(b) happening exactly one of the events is m + n (known as addition principle).
Example : There are 15 IITs in India and let each IIT has 10 branches, then the IITJEE topper can select the IIT
and branch in 15 × 10 = 150 number of ways.
Example : There are 15 IITs  & 20 NITs in India, then a student who cleared both IITJEE & AIEEE exams can
select an institute in (15 + 20) = 35 number of ways.
Illustration 1 : A college offers 6 courses in the morning and 4 in the evening. The possible number of choices
with the student if he wants to study one course in the morning and one in the evening is-
(A) 24 (B) 2 (C) 12 (D) 10
Solution : The student has 6 choices from the morning courses out of which he can select one course in
6 ways.
For the evening course, he has 4 choices out of which he can select one in 4 ways.
Hence the total number of ways 6 × 4 = 24. Ans.(A)
Illustration 2 : A college offers 6 courses in the morning and 4 in the evening. The number of ways a student
can select exactly one course, either in the morning or in the evening-
(A) 6 (B) 4 (C) 10 (D) 24
Solution : The student has 6 choices from the morning courses out of which he can select one course in
6 ways.
For the evening course, he has 4 choices out of which he can select one in 4 ways.
Hence the total number of ways 6 + 4 = 10. Ans. (C)
Do yourself - 1 :
(i) There are 3 ways to go from A to B, 2 ways to go from B to C and 1 way to go from A to C. In how many
ways  can a person travel from A to C ?
(ii) There are 2 red balls and 3 green balls. All balls are identical except colour. In how many ways can a
person  select two balls ?
2 . PERMUTATION & COMBINATION :
( a ) Factorial :  A Useful Notation : n! = n.(n – 1).(n – 2)..............3. 2. 1;
n! = n. (n – 1)! where n ? N
Note :
(i) 0! = 1! = 1
(ii) Factorials of negative integers are not defined.
(iii) n! is also denoted by n
PERMUTATION & COMBINATION
JEEMAIN.GURU
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N O D E 6 \ E \ D a t a \ 2 0 1 4 \ K o t a \ J E E - A d v a n c e d \ S M P \ M a t h s \ U n i t # 1 2 \ E N G \ 0 1 . P E R M U T A T I O N - C O M B I N A T I O N . p 6 5
J E E - M a t h e m a t i c s
1 . FUNDAMENTAL PRINCIPLE OF COUNTING  (counting without actual counting):
If an event A can occur in 'm' different ways and another event B can occur in 'n' different ways, then the total
number of different ways of-
(a) simultaneous occurrence of both events in a definite order is m× n. This can be extended to any number
of events (known as multiplication principle).
(b) happening exactly one of the events is m + n (known as addition principle).
Example : There are 15 IITs in India and let each IIT has 10 branches, then the IITJEE topper can select the IIT
and branch in 15 × 10 = 150 number of ways.
Example : There are 15 IITs  & 20 NITs in India, then a student who cleared both IITJEE & AIEEE exams can
select an institute in (15 + 20) = 35 number of ways.
Illustration 1 : A college offers 6 courses in the morning and 4 in the evening. The possible number of choices
with the student if he wants to study one course in the morning and one in the evening is-
(A) 24 (B) 2 (C) 12 (D) 10
Solution : The student has 6 choices from the morning courses out of which he can select one course in
6 ways.
For the evening course, he has 4 choices out of which he can select one in 4 ways.
Hence the total number of ways 6 × 4 = 24. Ans.(A)
Illustration 2 : A college offers 6 courses in the morning and 4 in the evening. The number of ways a student
can select exactly one course, either in the morning or in the evening-
(A) 6 (B) 4 (C) 10 (D) 24
Solution : The student has 6 choices from the morning courses out of which he can select one course in
6 ways.
For the evening course, he has 4 choices out of which he can select one in 4 ways.
Hence the total number of ways 6 + 4 = 10. Ans. (C)
Do yourself - 1 :
(i) There are 3 ways to go from A to B, 2 ways to go from B to C and 1 way to go from A to C. In how many
ways  can a person travel from A to C ?
(ii) There are 2 red balls and 3 green balls. All balls are identical except colour. In how many ways can a
person  select two balls ?
2 . PERMUTATION & COMBINATION :
( a ) Factorial :  A Useful Notation : n! = n.(n – 1).(n – 2)..............3. 2. 1;
n! = n. (n – 1)! where n ? N
Note :
(i) 0! = 1! = 1
(ii) Factorials of negative integers are not defined.
(iii) n! is also denoted by n
PERMUTATION & COMBINATION
JEEMAIN.GURU
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N O D E 6 \ E \ D a t a \ 2 0 1 4 \ K o t a \ J E E - A d v a n c e d \ S M P \ M a t h s \ U n i t # 1 2 \ E N G \ 0 1 . P E R M U T A T I O N - C O M B I N A T I O N . p 6 5
J E E - M a t h e m a t i c s
(iv) (2n)! =  2
n
.n! [1. 3. 5. 7........(2n – 1)]
(v) Prime factorisation of n! :  Let p be a prime number and n be a positive integer, then exponent of
p in n! is denoted by E
p 
(n!) and is given by
E
p
(n!) = 
n
p
? ?
? ?
? ?
 + 
2
n
p
? ?
? ?
? ?
 + 
3
n
p
? ?
? ?
? ?
 + ..... + 
k
n
p
? ?
? ?
? ?
where p
k
 < n < p
k+1 
 and [x] denotes the integral part of x.
If we isolate the power of each prime contained in any number n, then n can be written as
n = 
1
2
?
·
2
3
?
·
3
5
?
·
4
7
?
 .... where ?
i 
are whole numbers.
(b ) Permutation : Each of the arrangements in a definite order which can be made by taking some or all of
the things  at a time is called a PERMUTATION. In permutation, order of appearance of things is taken
into account; when the order is changed, a different permutation is obtained.
Generally, it involves the problems of  arrangements (standing in a line, seated in a row), problems on
digit,   problems on letters from a word etc.
n
P
r
 denotes the number of permutations of n different things, taken r at a time (n ? N, r ? W, r ? n)
n
P
r
  = n (n – 1) (n – 2) ............. (n – r + 1) = 
n!
(n r)! ?
Note :
(i)
n
P
n
 
= n!, 
   n
P
0
= 1, 
   n
P
1
= n
(ii) Number of arrangements of n distinct things taken all at a time = n!
(iii)
n
P
r
 
is also denoted by 
n
r
A or P(n,r).
(c) Combination :
Each of the groups or selections which can be made by taking some or all of the things without considering
the order of the things in each group is called a  COMBINATION.
Generally, involves the problem of  selections, choosing, distributed groups formation, committee formation,
geometrical problems etc.
n
C
r
 denotes the number of combinations of n different things taken r at a time (n ? N, r ? W, r < n)
n
r
n!
C
r!(n r)!
?
?
Note :
(i)
n
C
r
 is also denoted by 
n
r
? ?
? ?
? ?
 or C (n, r).
(ii)
n
P
r
 = 
n
C
r
. r!
Illustration 3: Find the exponent of 6 in 50!
So lu ti on: 2
50 50 50 50 50 50
E (50!)
2 4 8 16 32 64
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ?
 (where [  ] denotes integral part)
E
2
(50!) = 25 + 12 + 6 + 3 + 1 + 0 = 47
E
3
(50!) = 
50 50 50 50
3 9 27 81
? ? ? ? ? ? ? ?
? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
E
3
(50!) = 16 + 5 + 1 + 0 = 22
?    50! can be written as  50! = 2
47
. 3
22
.........
Therefore exponent of 6 in 50! = 22 Ans.
JEEMAIN.GURU
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N O D E 6 \ E \ D a t a \ 2 0 1 4 \ K o t a \ J E E - A d v a n c e d \ S M P \ M a t h s \ U n i t # 1 2 \ E N G \ 0 1 . P E R M U T A T I O N - C O M B I N A T I O N . p 6 5
J E E - M a t h e m a t i c s
1 . FUNDAMENTAL PRINCIPLE OF COUNTING  (counting without actual counting):
If an event A can occur in 'm' different ways and another event B can occur in 'n' different ways, then the total
number of different ways of-
(a) simultaneous occurrence of both events in a definite order is m× n. This can be extended to any number
of events (known as multiplication principle).
(b) happening exactly one of the events is m + n (known as addition principle).
Example : There are 15 IITs in India and let each IIT has 10 branches, then the IITJEE topper can select the IIT
and branch in 15 × 10 = 150 number of ways.
Example : There are 15 IITs  & 20 NITs in India, then a student who cleared both IITJEE & AIEEE exams can
select an institute in (15 + 20) = 35 number of ways.
Illustration 1 : A college offers 6 courses in the morning and 4 in the evening. The possible number of choices
with the student if he wants to study one course in the morning and one in the evening is-
(A) 24 (B) 2 (C) 12 (D) 10
Solution : The student has 6 choices from the morning courses out of which he can select one course in
6 ways.
For the evening course, he has 4 choices out of which he can select one in 4 ways.
Hence the total number of ways 6 × 4 = 24. Ans.(A)
Illustration 2 : A college offers 6 courses in the morning and 4 in the evening. The number of ways a student
can select exactly one course, either in the morning or in the evening-
(A) 6 (B) 4 (C) 10 (D) 24
Solution : The student has 6 choices from the morning courses out of which he can select one course in
6 ways.
For the evening course, he has 4 choices out of which he can select one in 4 ways.
Hence the total number of ways 6 + 4 = 10. Ans. (C)
Do yourself - 1 :
(i) There are 3 ways to go from A to B, 2 ways to go from B to C and 1 way to go from A to C. In how many
ways  can a person travel from A to C ?
(ii) There are 2 red balls and 3 green balls. All balls are identical except colour. In how many ways can a
person  select two balls ?
2 . PERMUTATION & COMBINATION :
( a ) Factorial :  A Useful Notation : n! = n.(n – 1).(n – 2)..............3. 2. 1;
n! = n. (n – 1)! where n ? N
Note :
(i) 0! = 1! = 1
(ii) Factorials of negative integers are not defined.
(iii) n! is also denoted by n
PERMUTATION & COMBINATION
JEEMAIN.GURU
2
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N O D E 6 \ E \ D a t a \ 2 0 1 4 \ K o t a \ J E E - A d v a n c e d \ S M P \ M a t h s \ U n i t # 1 2 \ E N G \ 0 1 . P E R M U T A T I O N - C O M B I N A T I O N . p 6 5
J E E - M a t h e m a t i c s
(iv) (2n)! =  2
n
.n! [1. 3. 5. 7........(2n – 1)]
(v) Prime factorisation of n! :  Let p be a prime number and n be a positive integer, then exponent of
p in n! is denoted by E
p 
(n!) and is given by
E
p
(n!) = 
n
p
? ?
? ?
? ?
 + 
2
n
p
? ?
? ?
? ?
 + 
3
n
p
? ?
? ?
? ?
 + ..... + 
k
n
p
? ?
? ?
? ?
where p
k
 < n < p
k+1 
 and [x] denotes the integral part of x.
If we isolate the power of each prime contained in any number n, then n can be written as
n = 
1
2
?
·
2
3
?
·
3
5
?
·
4
7
?
 .... where ?
i 
are whole numbers.
(b ) Permutation : Each of the arrangements in a definite order which can be made by taking some or all of
the things  at a time is called a PERMUTATION. In permutation, order of appearance of things is taken
into account; when the order is changed, a different permutation is obtained.
Generally, it involves the problems of  arrangements (standing in a line, seated in a row), problems on
digit,   problems on letters from a word etc.
n
P
r
 denotes the number of permutations of n different things, taken r at a time (n ? N, r ? W, r ? n)
n
P
r
  = n (n – 1) (n – 2) ............. (n – r + 1) = 
n!
(n r)! ?
Note :
(i)
n
P
n
 
= n!, 
   n
P
0
= 1, 
   n
P
1
= n
(ii) Number of arrangements of n distinct things taken all at a time = n!
(iii)
n
P
r
 
is also denoted by 
n
r
A or P(n,r).
(c) Combination :
Each of the groups or selections which can be made by taking some or all of the things without considering
the order of the things in each group is called a  COMBINATION.
Generally, involves the problem of  selections, choosing, distributed groups formation, committee formation,
geometrical problems etc.
n
C
r
 denotes the number of combinations of n different things taken r at a time (n ? N, r ? W, r < n)
n
r
n!
C
r!(n r)!
?
?
Note :
(i)
n
C
r
 is also denoted by 
n
r
? ?
? ?
? ?
 or C (n, r).
(ii)
n
P
r
 = 
n
C
r
. r!
Illustration 3: Find the exponent of 6 in 50!
So lu ti on: 2
50 50 50 50 50 50
E (50!)
2 4 8 16 32 64
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ?
 (where [  ] denotes integral part)
E
2
(50!) = 25 + 12 + 6 + 3 + 1 + 0 = 47
E
3
(50!) = 
50 50 50 50
3 9 27 81
? ? ? ? ? ? ? ?
? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
E
3
(50!) = 16 + 5 + 1 + 0 = 22
?    50! can be written as  50! = 2
47
. 3
22
.........
Therefore exponent of 6 in 50! = 22 Ans.
JEEMAIN.GURU
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N O D E 6 \ E \ D a t a \ 2 0 1 4 \ K o t a \ J E E - A d v a n c e d \ S M P \ M a t h s \ U n i t # 1 2 \ E N G \ 0 1 . P E R M U T A T I O N - C O M B I N A T I O N . p 6 5
J E E - M a t h e m a t i c s
Illustration 4 : If a denotes the number of permutations of (x + 2) things taken all at a time, b the number of
permutations of x things taken 11 at a time and c the number of permutations of (x – 11) things
taken all at a time such that a = 182 bc, then the value of x is
(A) 15 (B) 12 (C) 10 (D) 18
Solution : ? ?
x 2
x 2
P a a x 2 !
?
?
? ? ? ?
? ?
x
11
x!
P b b
x 11 !
? ? ?
?
and ? ?
x 11
x 11
P c c x 11 !
?
?
? ? ? ?
a 182bc ? ?
? ?
? ?
? ?
x!
x 2 ! 182 x 11 !
x 11 !
? ? ?
?
 ? ? ? ? x 2 x 1 182 14 13 ? ? ? ? ? ?
x 1 13 x 12 ? ? ? ? ? Ans. (B)
Illustration 5 : A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be drawn
so that there are atleast two balls of each colour ?
Solution : The selections of 6 balls, consisting of atleast two balls of each colour from 5 red and 6 white balls,
can be made in the following ways
5 6
2 4
5 6
3 3
5 6
4 2
Red balls (5) White balls (6) Number of ways
2 4 C C 150
3 3 C C 200
4 2 C C 75
? ?
? ?
? ?
Therefore total number of ways = 425 Ans.
Illustration 6 : How many 4 letter words can be formed from the letters of the word 'ANSWER' ? How many of
these words start with a vowel ?
Solution : Number of ways of arranging 4 different letters from 6 different letters are 
6
4
6!
C 4! 360
2!
? ?
.
There are two vowels (A & E) in the word 'ANSWER'.
Total number of 4 letter words starting with A : A _ _ _ = 
5
3
5!
C 3! 60
2!
? ?
Total number of 4 letter words starting with E : E _ _ _  = 
5
3
5!
C 3! 60
2!
? ?
? Total number of 4 letter words starting with a vowel = 60 + 60 = 120. Ans.
Illustration 7 : If all the letters of the word 'RAPID' are arranged in all possible manner as they are in a dictionary,
then find the rank of the word 'RAPID'.
Solution : First of all, arrange all letters of given word alphabetically : 'ADIPR'
Total number of words starting with A _ _ _ _ = 4! = 24
Total number of words starting with D _ _ _ _ = 4! = 24
Total number of words starting with I _ _ _ _ = 4! = 24
Total number of words starting with P _ _ _ _ = 4! = 24
Total number of words starting with RAD _ _ = 2! = 2
Total number of words starting with RAI _ _ = 2! = 2
Total number of words starting with RAPD _          = 1
Total number of words starting with RAPI _      = 1
? Rank of the word RAPID = 24 + 24 + 24 + 24 + 2 + 2 + 1 + 1 = 102 Ans.
JEEMAIN.GURU
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N O D E 6 \ E \ D a t a \ 2 0 1 4 \ K o t a \ J E E - A d v a n c e d \ S M P \ M a t h s \ U n i t # 1 2 \ E N G \ 0 1 . P E R M U T A T I O N - C O M B I N A T I O N . p 6 5
J E E - M a t h e m a t i c s
1 . FUNDAMENTAL PRINCIPLE OF COUNTING  (counting without actual counting):
If an event A can occur in 'm' different ways and another event B can occur in 'n' different ways, then the total
number of different ways of-
(a) simultaneous occurrence of both events in a definite order is m× n. This can be extended to any number
of events (known as multiplication principle).
(b) happening exactly one of the events is m + n (known as addition principle).
Example : There are 15 IITs in India and let each IIT has 10 branches, then the IITJEE topper can select the IIT
and branch in 15 × 10 = 150 number of ways.
Example : There are 15 IITs  & 20 NITs in India, then a student who cleared both IITJEE & AIEEE exams can
select an institute in (15 + 20) = 35 number of ways.
Illustration 1 : A college offers 6 courses in the morning and 4 in the evening. The possible number of choices
with the student if he wants to study one course in the morning and one in the evening is-
(A) 24 (B) 2 (C) 12 (D) 10
Solution : The student has 6 choices from the morning courses out of which he can select one course in
6 ways.
For the evening course, he has 4 choices out of which he can select one in 4 ways.
Hence the total number of ways 6 × 4 = 24. Ans.(A)
Illustration 2 : A college offers 6 courses in the morning and 4 in the evening. The number of ways a student
can select exactly one course, either in the morning or in the evening-
(A) 6 (B) 4 (C) 10 (D) 24
Solution : The student has 6 choices from the morning courses out of which he can select one course in
6 ways.
For the evening course, he has 4 choices out of which he can select one in 4 ways.
Hence the total number of ways 6 + 4 = 10. Ans. (C)
Do yourself - 1 :
(i) There are 3 ways to go from A to B, 2 ways to go from B to C and 1 way to go from A to C. In how many
ways  can a person travel from A to C ?
(ii) There are 2 red balls and 3 green balls. All balls are identical except colour. In how many ways can a
person  select two balls ?
2 . PERMUTATION & COMBINATION :
( a ) Factorial :  A Useful Notation : n! = n.(n – 1).(n – 2)..............3. 2. 1;
n! = n. (n – 1)! where n ? N
Note :
(i) 0! = 1! = 1
(ii) Factorials of negative integers are not defined.
(iii) n! is also denoted by n
PERMUTATION & COMBINATION
JEEMAIN.GURU
2
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N O D E 6 \ E \ D a t a \ 2 0 1 4 \ K o t a \ J E E - A d v a n c e d \ S M P \ M a t h s \ U n i t # 1 2 \ E N G \ 0 1 . P E R M U T A T I O N - C O M B I N A T I O N . p 6 5
J E E - M a t h e m a t i c s
(iv) (2n)! =  2
n
.n! [1. 3. 5. 7........(2n – 1)]
(v) Prime factorisation of n! :  Let p be a prime number and n be a positive integer, then exponent of
p in n! is denoted by E
p 
(n!) and is given by
E
p
(n!) = 
n
p
? ?
? ?
? ?
 + 
2
n
p
? ?
? ?
? ?
 + 
3
n
p
? ?
? ?
? ?
 + ..... + 
k
n
p
? ?
? ?
? ?
where p
k
 < n < p
k+1 
 and [x] denotes the integral part of x.
If we isolate the power of each prime contained in any number n, then n can be written as
n = 
1
2
?
·
2
3
?
·
3
5
?
·
4
7
?
 .... where ?
i 
are whole numbers.
(b ) Permutation : Each of the arrangements in a definite order which can be made by taking some or all of
the things  at a time is called a PERMUTATION. In permutation, order of appearance of things is taken
into account; when the order is changed, a different permutation is obtained.
Generally, it involves the problems of  arrangements (standing in a line, seated in a row), problems on
digit,   problems on letters from a word etc.
n
P
r
 denotes the number of permutations of n different things, taken r at a time (n ? N, r ? W, r ? n)
n
P
r
  = n (n – 1) (n – 2) ............. (n – r + 1) = 
n!
(n r)! ?
Note :
(i)
n
P
n
 
= n!, 
   n
P
0
= 1, 
   n
P
1
= n
(ii) Number of arrangements of n distinct things taken all at a time = n!
(iii)
n
P
r
 
is also denoted by 
n
r
A or P(n,r).
(c) Combination :
Each of the groups or selections which can be made by taking some or all of the things without considering
the order of the things in each group is called a  COMBINATION.
Generally, involves the problem of  selections, choosing, distributed groups formation, committee formation,
geometrical problems etc.
n
C
r
 denotes the number of combinations of n different things taken r at a time (n ? N, r ? W, r < n)
n
r
n!
C
r!(n r)!
?
?
Note :
(i)
n
C
r
 is also denoted by 
n
r
? ?
? ?
? ?
 or C (n, r).
(ii)
n
P
r
 = 
n
C
r
. r!
Illustration 3: Find the exponent of 6 in 50!
So lu ti on: 2
50 50 50 50 50 50
E (50!)
2 4 8 16 32 64
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ?
 (where [  ] denotes integral part)
E
2
(50!) = 25 + 12 + 6 + 3 + 1 + 0 = 47
E
3
(50!) = 
50 50 50 50
3 9 27 81
? ? ? ? ? ? ? ?
? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
E
3
(50!) = 16 + 5 + 1 + 0 = 22
?    50! can be written as  50! = 2
47
. 3
22
.........
Therefore exponent of 6 in 50! = 22 Ans.
JEEMAIN.GURU
E
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N O D E 6 \ E \ D a t a \ 2 0 1 4 \ K o t a \ J E E - A d v a n c e d \ S M P \ M a t h s \ U n i t # 1 2 \ E N G \ 0 1 . P E R M U T A T I O N - C O M B I N A T I O N . p 6 5
J E E - M a t h e m a t i c s
Illustration 4 : If a denotes the number of permutations of (x + 2) things taken all at a time, b the number of
permutations of x things taken 11 at a time and c the number of permutations of (x – 11) things
taken all at a time such that a = 182 bc, then the value of x is
(A) 15 (B) 12 (C) 10 (D) 18
Solution : ? ?
x 2
x 2
P a a x 2 !
?
?
? ? ? ?
? ?
x
11
x!
P b b
x 11 !
? ? ?
?
and ? ?
x 11
x 11
P c c x 11 !
?
?
? ? ? ?
a 182bc ? ?
? ?
? ?
? ?
x!
x 2 ! 182 x 11 !
x 11 !
? ? ?
?
 ? ? ? ? x 2 x 1 182 14 13 ? ? ? ? ? ?
x 1 13 x 12 ? ? ? ? ? Ans. (B)
Illustration 5 : A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be drawn
so that there are atleast two balls of each colour ?
Solution : The selections of 6 balls, consisting of atleast two balls of each colour from 5 red and 6 white balls,
can be made in the following ways
5 6
2 4
5 6
3 3
5 6
4 2
Red balls (5) White balls (6) Number of ways
2 4 C C 150
3 3 C C 200
4 2 C C 75
? ?
? ?
? ?
Therefore total number of ways = 425 Ans.
Illustration 6 : How many 4 letter words can be formed from the letters of the word 'ANSWER' ? How many of
these words start with a vowel ?
Solution : Number of ways of arranging 4 different letters from 6 different letters are 
6
4
6!
C 4! 360
2!
? ?
.
There are two vowels (A & E) in the word 'ANSWER'.
Total number of 4 letter words starting with A : A _ _ _ = 
5
3
5!
C 3! 60
2!
? ?
Total number of 4 letter words starting with E : E _ _ _  = 
5
3
5!
C 3! 60
2!
? ?
? Total number of 4 letter words starting with a vowel = 60 + 60 = 120. Ans.
Illustration 7 : If all the letters of the word 'RAPID' are arranged in all possible manner as they are in a dictionary,
then find the rank of the word 'RAPID'.
Solution : First of all, arrange all letters of given word alphabetically : 'ADIPR'
Total number of words starting with A _ _ _ _ = 4! = 24
Total number of words starting with D _ _ _ _ = 4! = 24
Total number of words starting with I _ _ _ _ = 4! = 24
Total number of words starting with P _ _ _ _ = 4! = 24
Total number of words starting with RAD _ _ = 2! = 2
Total number of words starting with RAI _ _ = 2! = 2
Total number of words starting with RAPD _          = 1
Total number of words starting with RAPI _      = 1
? Rank of the word RAPID = 24 + 24 + 24 + 24 + 2 + 2 + 1 + 1 = 102 Ans.
JEEMAIN.GURU
4
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J E E - M a t h e m a t i c s
Do yourself -2 :
(i) Find the exponent of 10 in 
75
C
25
.
(ii) If 
10
P
r
 
= 5040, then find the value of r.
(iii) Find the number of ways of selecting 4 even numbers from the set of first 100 natural numbers.
(iv) If all letters of the word 'RANK' are arranged in all possible manner as they are in a dictionary, then
find the rank of the word 'RANK'.
(v) How many words can be formed using all letters of the word 'LEARN' ? In how many of these words
vowels are together ?
3 . PROPERTIES OF 
n
P
r 
 and 
n
C
r
 :
( a ) The number of permutation of n different objects taken r at a time, when p particular objects are  always
to be included is  r!.
n–p
C
r–p
 (p ? r ? n)
(b ) The number of permutations of n different objects taken r at a time, when repetition is allowed any
number of times is  n
r
.
(c) Following  properties of 
n
C
r
 should be remembered :
(i)
 n
C
r
 
= 
n
C
n–r
 ; 
n
C
0
 
= 
n
C
n
 = 1 (ii)
n
C
x
 
= 
n
C
y
 ? x = y or  x + y = n
(iii)
n
C
r
 
+ 
n
C
r–1
 = 
n+1
C
r
(iv)
n
C
0
 
+
 n
C
1
 + 
n
C
2 
+ ............ + 
n
C
n
 = 2
n
(v)
n
C
r
 = 
n
r
 n–1
C
r–1
(vi)
n
C
r
 is maximum when  
n
r
2
? if n is even  &  
n 1
r
2
?
? or 
n 1
r
2
?
? if n is odd.
(d ) The number of combinations of n different things taking r at a time,
(i) when p particular things are always to be included = 
n – p
C
r–p
(ii) when p particular things are always to be excluded = 
n – p
C
r
(iii) when p particular things are always to be included and q particular things are to be excluded
? 
n – p – q
C
r–p
Illustration 8 : There are 6 pockets in the coat of a person. In how many ways can he put 4 pens in these
pockets?
(A) 360 (B) 1296 (C) 4096 (D) none of these
Solution : First pen can be put in 6 ways.
Similarly each of second, third and fourth pen can be put in 6 ways.
Hence total number of ways = 6 × 6 × 6 × 6 = 1296 Ans.(B)
Illustration 9 : A delegation of four students is to be selected from a total of 12 students. In how many ways can
the delegation be selected, if-
(a) all the students are equally willing ?
(b) two particular students have to be included in the delegation ?
(c) two particular students do not wish to be together in the delegation ?
(d) two particular students wish to be included together only ?
(e) two particular students refuse to be together and two other particular students wish to be
together only in the delegation ?
Solution : (a) Formation of delegation means selection of 4 out of 12.
Hence the number of ways = 
12
C
4
 = 495.
(b) If two particular students are already selected. Here we need to select only 2 out of the
remaining 10. Hence the number of ways = 
10
C
2
 = 45.
(c) The number of ways in which both are selected = 45. Hence the number of ways in which
the two are not included together = 495 – 45 = 450
(d) There are two possible cases
(i)   Either both are selected. In this case, the number of ways in which the selection can be
made = 45.
(ii)  Or both are not selected. In this case all the four students are selected from the remaining
ten students. This can be done in 
10
C
4
 = 210 ways.
Hence the total number of ways of selection = 45 + 210 = 255
JEEMAIN.GURU
Page 5


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J E E - M a t h e m a t i c s
1 . FUNDAMENTAL PRINCIPLE OF COUNTING  (counting without actual counting):
If an event A can occur in 'm' different ways and another event B can occur in 'n' different ways, then the total
number of different ways of-
(a) simultaneous occurrence of both events in a definite order is m× n. This can be extended to any number
of events (known as multiplication principle).
(b) happening exactly one of the events is m + n (known as addition principle).
Example : There are 15 IITs in India and let each IIT has 10 branches, then the IITJEE topper can select the IIT
and branch in 15 × 10 = 150 number of ways.
Example : There are 15 IITs  & 20 NITs in India, then a student who cleared both IITJEE & AIEEE exams can
select an institute in (15 + 20) = 35 number of ways.
Illustration 1 : A college offers 6 courses in the morning and 4 in the evening. The possible number of choices
with the student if he wants to study one course in the morning and one in the evening is-
(A) 24 (B) 2 (C) 12 (D) 10
Solution : The student has 6 choices from the morning courses out of which he can select one course in
6 ways.
For the evening course, he has 4 choices out of which he can select one in 4 ways.
Hence the total number of ways 6 × 4 = 24. Ans.(A)
Illustration 2 : A college offers 6 courses in the morning and 4 in the evening. The number of ways a student
can select exactly one course, either in the morning or in the evening-
(A) 6 (B) 4 (C) 10 (D) 24
Solution : The student has 6 choices from the morning courses out of which he can select one course in
6 ways.
For the evening course, he has 4 choices out of which he can select one in 4 ways.
Hence the total number of ways 6 + 4 = 10. Ans. (C)
Do yourself - 1 :
(i) There are 3 ways to go from A to B, 2 ways to go from B to C and 1 way to go from A to C. In how many
ways  can a person travel from A to C ?
(ii) There are 2 red balls and 3 green balls. All balls are identical except colour. In how many ways can a
person  select two balls ?
2 . PERMUTATION & COMBINATION :
( a ) Factorial :  A Useful Notation : n! = n.(n – 1).(n – 2)..............3. 2. 1;
n! = n. (n – 1)! where n ? N
Note :
(i) 0! = 1! = 1
(ii) Factorials of negative integers are not defined.
(iii) n! is also denoted by n
PERMUTATION & COMBINATION
JEEMAIN.GURU
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J E E - M a t h e m a t i c s
(iv) (2n)! =  2
n
.n! [1. 3. 5. 7........(2n – 1)]
(v) Prime factorisation of n! :  Let p be a prime number and n be a positive integer, then exponent of
p in n! is denoted by E
p 
(n!) and is given by
E
p
(n!) = 
n
p
? ?
? ?
? ?
 + 
2
n
p
? ?
? ?
? ?
 + 
3
n
p
? ?
? ?
? ?
 + ..... + 
k
n
p
? ?
? ?
? ?
where p
k
 < n < p
k+1 
 and [x] denotes the integral part of x.
If we isolate the power of each prime contained in any number n, then n can be written as
n = 
1
2
?
·
2
3
?
·
3
5
?
·
4
7
?
 .... where ?
i 
are whole numbers.
(b ) Permutation : Each of the arrangements in a definite order which can be made by taking some or all of
the things  at a time is called a PERMUTATION. In permutation, order of appearance of things is taken
into account; when the order is changed, a different permutation is obtained.
Generally, it involves the problems of  arrangements (standing in a line, seated in a row), problems on
digit,   problems on letters from a word etc.
n
P
r
 denotes the number of permutations of n different things, taken r at a time (n ? N, r ? W, r ? n)
n
P
r
  = n (n – 1) (n – 2) ............. (n – r + 1) = 
n!
(n r)! ?
Note :
(i)
n
P
n
 
= n!, 
   n
P
0
= 1, 
   n
P
1
= n
(ii) Number of arrangements of n distinct things taken all at a time = n!
(iii)
n
P
r
 
is also denoted by 
n
r
A or P(n,r).
(c) Combination :
Each of the groups or selections which can be made by taking some or all of the things without considering
the order of the things in each group is called a  COMBINATION.
Generally, involves the problem of  selections, choosing, distributed groups formation, committee formation,
geometrical problems etc.
n
C
r
 denotes the number of combinations of n different things taken r at a time (n ? N, r ? W, r < n)
n
r
n!
C
r!(n r)!
?
?
Note :
(i)
n
C
r
 is also denoted by 
n
r
? ?
? ?
? ?
 or C (n, r).
(ii)
n
P
r
 = 
n
C
r
. r!
Illustration 3: Find the exponent of 6 in 50!
So lu ti on: 2
50 50 50 50 50 50
E (50!)
2 4 8 16 32 64
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ?
 (where [  ] denotes integral part)
E
2
(50!) = 25 + 12 + 6 + 3 + 1 + 0 = 47
E
3
(50!) = 
50 50 50 50
3 9 27 81
? ? ? ? ? ? ? ?
? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
E
3
(50!) = 16 + 5 + 1 + 0 = 22
?    50! can be written as  50! = 2
47
. 3
22
.........
Therefore exponent of 6 in 50! = 22 Ans.
JEEMAIN.GURU
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J E E - M a t h e m a t i c s
Illustration 4 : If a denotes the number of permutations of (x + 2) things taken all at a time, b the number of
permutations of x things taken 11 at a time and c the number of permutations of (x – 11) things
taken all at a time such that a = 182 bc, then the value of x is
(A) 15 (B) 12 (C) 10 (D) 18
Solution : ? ?
x 2
x 2
P a a x 2 !
?
?
? ? ? ?
? ?
x
11
x!
P b b
x 11 !
? ? ?
?
and ? ?
x 11
x 11
P c c x 11 !
?
?
? ? ? ?
a 182bc ? ?
? ?
? ?
? ?
x!
x 2 ! 182 x 11 !
x 11 !
? ? ?
?
 ? ? ? ? x 2 x 1 182 14 13 ? ? ? ? ? ?
x 1 13 x 12 ? ? ? ? ? Ans. (B)
Illustration 5 : A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be drawn
so that there are atleast two balls of each colour ?
Solution : The selections of 6 balls, consisting of atleast two balls of each colour from 5 red and 6 white balls,
can be made in the following ways
5 6
2 4
5 6
3 3
5 6
4 2
Red balls (5) White balls (6) Number of ways
2 4 C C 150
3 3 C C 200
4 2 C C 75
? ?
? ?
? ?
Therefore total number of ways = 425 Ans.
Illustration 6 : How many 4 letter words can be formed from the letters of the word 'ANSWER' ? How many of
these words start with a vowel ?
Solution : Number of ways of arranging 4 different letters from 6 different letters are 
6
4
6!
C 4! 360
2!
? ?
.
There are two vowels (A & E) in the word 'ANSWER'.
Total number of 4 letter words starting with A : A _ _ _ = 
5
3
5!
C 3! 60
2!
? ?
Total number of 4 letter words starting with E : E _ _ _  = 
5
3
5!
C 3! 60
2!
? ?
? Total number of 4 letter words starting with a vowel = 60 + 60 = 120. Ans.
Illustration 7 : If all the letters of the word 'RAPID' are arranged in all possible manner as they are in a dictionary,
then find the rank of the word 'RAPID'.
Solution : First of all, arrange all letters of given word alphabetically : 'ADIPR'
Total number of words starting with A _ _ _ _ = 4! = 24
Total number of words starting with D _ _ _ _ = 4! = 24
Total number of words starting with I _ _ _ _ = 4! = 24
Total number of words starting with P _ _ _ _ = 4! = 24
Total number of words starting with RAD _ _ = 2! = 2
Total number of words starting with RAI _ _ = 2! = 2
Total number of words starting with RAPD _          = 1
Total number of words starting with RAPI _      = 1
? Rank of the word RAPID = 24 + 24 + 24 + 24 + 2 + 2 + 1 + 1 = 102 Ans.
JEEMAIN.GURU
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J E E - M a t h e m a t i c s
Do yourself -2 :
(i) Find the exponent of 10 in 
75
C
25
.
(ii) If 
10
P
r
 
= 5040, then find the value of r.
(iii) Find the number of ways of selecting 4 even numbers from the set of first 100 natural numbers.
(iv) If all letters of the word 'RANK' are arranged in all possible manner as they are in a dictionary, then
find the rank of the word 'RANK'.
(v) How many words can be formed using all letters of the word 'LEARN' ? In how many of these words
vowels are together ?
3 . PROPERTIES OF 
n
P
r 
 and 
n
C
r
 :
( a ) The number of permutation of n different objects taken r at a time, when p particular objects are  always
to be included is  r!.
n–p
C
r–p
 (p ? r ? n)
(b ) The number of permutations of n different objects taken r at a time, when repetition is allowed any
number of times is  n
r
.
(c) Following  properties of 
n
C
r
 should be remembered :
(i)
 n
C
r
 
= 
n
C
n–r
 ; 
n
C
0
 
= 
n
C
n
 = 1 (ii)
n
C
x
 
= 
n
C
y
 ? x = y or  x + y = n
(iii)
n
C
r
 
+ 
n
C
r–1
 = 
n+1
C
r
(iv)
n
C
0
 
+
 n
C
1
 + 
n
C
2 
+ ............ + 
n
C
n
 = 2
n
(v)
n
C
r
 = 
n
r
 n–1
C
r–1
(vi)
n
C
r
 is maximum when  
n
r
2
? if n is even  &  
n 1
r
2
?
? or 
n 1
r
2
?
? if n is odd.
(d ) The number of combinations of n different things taking r at a time,
(i) when p particular things are always to be included = 
n – p
C
r–p
(ii) when p particular things are always to be excluded = 
n – p
C
r
(iii) when p particular things are always to be included and q particular things are to be excluded
? 
n – p – q
C
r–p
Illustration 8 : There are 6 pockets in the coat of a person. In how many ways can he put 4 pens in these
pockets?
(A) 360 (B) 1296 (C) 4096 (D) none of these
Solution : First pen can be put in 6 ways.
Similarly each of second, third and fourth pen can be put in 6 ways.
Hence total number of ways = 6 × 6 × 6 × 6 = 1296 Ans.(B)
Illustration 9 : A delegation of four students is to be selected from a total of 12 students. In how many ways can
the delegation be selected, if-
(a) all the students are equally willing ?
(b) two particular students have to be included in the delegation ?
(c) two particular students do not wish to be together in the delegation ?
(d) two particular students wish to be included together only ?
(e) two particular students refuse to be together and two other particular students wish to be
together only in the delegation ?
Solution : (a) Formation of delegation means selection of 4 out of 12.
Hence the number of ways = 
12
C
4
 = 495.
(b) If two particular students are already selected. Here we need to select only 2 out of the
remaining 10. Hence the number of ways = 
10
C
2
 = 45.
(c) The number of ways in which both are selected = 45. Hence the number of ways in which
the two are not included together = 495 – 45 = 450
(d) There are two possible cases
(i)   Either both are selected. In this case, the number of ways in which the selection can be
made = 45.
(ii)  Or both are not selected. In this case all the four students are selected from the remaining
ten students. This can be done in 
10
C
4
 = 210 ways.
Hence the total number of ways of selection = 45 + 210 = 255
JEEMAIN.GURU
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J E E - M a t h e m a t i c s
(e) We assume that students A and B wish to be selected together and students C and D do not
wish to be together. Now there are following 6 cases.
(i) (A, B, C) selected, (D) not selected
(ii) (A, B, D) selected, (C) not selected
(iii) (A, B) selected, (C, D) not selected
(iv) (C)  selected, (A, B, D) not selected
(v) (D)  selected, (A, B, C) not selected
(vi) A, B, C, D  not selected
For (i) the number of ways of selection = 
8
C
1
 = 8
For (ii) the number of ways of selection = 
8
C
1
 = 8
For (iii) the number of ways of selection = 
8
C
2
 = 28
For (iv) the number of ways of selection = 
8
C
3
 = 56
For (v) the number of ways of selection = 
8
C
3
 = 56
For (vi) the number of ways of selection = 
8
C
4
 = 70
Hence total number of ways = 8 + 8 + 28 + 56 + 56 + 70 = 226. Ans.
Illustration 10: In the given figure of squares, 6 A's should be written in such a
manner that every row contains at least one 'A'. In how many
number of ways is it possible ?                           
(A) 24 (B) 25 (C) 26 (D) 27
Solution : There are 8 squares and 6 'A' in given figure. First we can put 6 'A' in these 8 squares by 
8
C
6
number of ways.
(I)   A
A
A A A
A
(II)   A
A
A A A
A
According to question, atleast one 'A' should be included in each row. So after subtracting these
two cases, number of ways are = (
8
C
6
 – 2) = 28 – 2 = 26.      Ans. (C)
Illustration 11: There are three coplanar parallel lines. If any p points are taken on each of the lines, the maximum
number of triangles with vertices at these points is :
(A) 3p
2
 (p – 1) + 1 (B) 3p
2
 (p – 1) (C) p
2
 (4p – 3) (D) none of these
Solution : The number of triangles with vertices on different lines = 
p
C
1
 × 
p
C
1
 × 
p
C
1
 = p
3
The number of triangles with two vertices on one line and the third vertex on any one of the other
two lines = 
3
C
1
 {
p
C
2
 × 
2p
C
1
} = 6p.
p(p 1)
2
?
So, the required number of triangles = p
3
 + 3p
2
 (p – 1) = p
2 
(4p – 3) Ans. (C)
Illustration 12: There are 10 points in a row. In how many ways can 4 points be selected such that no two of them
are consecutive ?
Solution : Total number of remaining non-selected points = 6
. . . . . .
Total number of gaps made by these 6 points = 6 + 1 = 7
If we select 4 gaps out of these 7 gaps and put 4 points in selected  gaps then the new points will
represent 4 points such that no two of them are consecutive.
x . . x . x . . x .
Total number of ways of  selecting 4 gaps out of 7 gaps = 
7
C
4
Ans.
In general, total number of ways of selection of r points out of n points in a row such
that no two of them are consecutive : 
n–r+1
C
r
JEEMAIN.GURU
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