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PERMUTATION AND COMBINATIONS
Book for Boards, NTSE, IIT JEE- Mains/Advanced, CAT, Olympiad & Software Engineers
i
ii
Contents
Contents
PERMUTATION AND COMBINATIONS
Book for Boards, NTSE, IIT JEE- Mains/Advanced, CAT, Olympiad & Software Engineers
Ramesh Chandra
B.Tech IIT Kanpur (Mechanical Engineering)
iii
iv
Contents
Notion Press
5 Muthu Kalathy Street, Triplicane,
Chennai - 600 005, India
First Published by Notion Press 2014
Copyright © Ramesh Chandra 2014
All Rights Reserved.
ISBN: 978-93-84391-47-8
This book is sold subject to condition that it shall not by way of trade or otherwise, be lent, resold or hired out, circulated
and no reproduction in any form, in whole or in part (except for brief quotations in critical articles or reviews) may be
made without written permission of the publishers.
This book has been published in good faith that the work of the author is original. All efforts have been taken to make the
material error-free. However, the author and the publisher disclaim the responsibility for any inadvertent errors.
Contents
CONTENTS
Page No
1.
Basic requirement to the topic
(a) Factorial (!)
(b) Summation
(c) Exercise - I (Subjective) + (Answer Key)
(d) Binomial coefficients ( n C r ) and some important formulae’s
2
2
4
6
7
(e) Solved examples to explore above formulae’s
8
2.
Counting
(a) Addition principle
(b) Multiplication principle
(c) Palindromes
(d) Solved problems to explore counting
(e) Inclusion – Exclusion Principle
(f) Exercise II (Subjective) + (Answer Key)
(g) Exercise - I (objective) + (Answer key)
11
11
12
14
15
25
27
29
3.
Permutations and Combinations
(a) Combinations (Selections)
(b) Permutations (Arrangement)
(i) Permutation when all the objects are distinct.
(ii) Permutations when some object are identical.
(c) Restricted selection / arrangement
(d) Other possible selections combinations when all the objects are not distinct
(e) Combinations ehen all the objects are not distinct
(f) Exercise – II (objective) + (Answer key)
Building block – TEST
30
30
34
36
42
45
49
50
53
54
4.
Circular permutations
(a) Introduction
(b) Garlands
58
58
61
5.
Number Theory
(a) Exponent of prime numbers in n!
(b) Number of divisors of a given natural number (N)
(c) Sum of divisor of natural number N
(d) Derangements
64
64
66
68
69
6.
Grouping
(a) Division and distribution of objects
(b) Multinomial theorem
(c) Results of binomial theorem
(d) Mathematical modeling
72
72
78
79
80
7.
Functional mapping
96

v
vi
Contents
8.
C.B.S.E board - N.C.E.R.T – XI (Mathematics Problems)
(a) Exercise 7.1
(b) Exercise 7.2
(c) Exercise 7.3
(d) Exercise 7.4
(e) Miscellaneous exercise on chapter 7
(f) C.B.S.E board - N.C.E.R.T – XI - Mathematics (answer key)
(g) Hints / Solutions
C.B.S.E board - N.C.E.R.T – XI (Mathematics)
102
102
102
102
103
104
104
105
105
9.
H.S.C. Board – XI CLASS (Mathematics Problems)
(a) Exercise 5.1
(b) Exercise 5.2
(c) Exercise 5.3
(d) Exercise 5.4
(e) Exercise 5.5
(f) Exercise 5.6
(g) Exercise: 5.7
(h) Miscellaneous exercise – 5
(i) H.S.C. BOARD – XI CLASS - Mathematics (answer key)
(j) Hints / solution
H.S.C. BOARD – XI Class (Mathematics)
110
110
111
112
113
113
114
115
115
117
119
119
10.
ENGINEERING EXAM PROBLEMS
(a) Objective Problems (Engineering Exam Archive) + (Answer key)
(b) IITJEE – Advanced
(i) Exercise–I
(ii) Exercise–II
(c) Answer Key
(i) Building block – TEST (class 8th, 9th, 10th)
(ii) Subjective Problems – IIT- JEE /Advanced
(iii) Exercise – I
(iv) Exercise– II
138
138
146
146
148
151
151
151
151
152
11.
ARCHIVE
(a) Archive – IIT-JEE – Mains/AIEEE
(b) Archive – IIT- JEE – Advanced (Objective + Subjective)
(c) Archive - CAT Exam
ANSWER – KEY
(a) Archive – IIT- JEE – Mains/AIEEE
(b) Archive – IIT- JEE – Advanced
(c) Archive - CAT Exam
SOLUTIONS
(a) Archive – IIT JEE – Advanced
(Objective Questions + Subjective Problems)
PUZZLE – SECTION
153
153
156
161
164
164
164
165
165
165

172
Contents
12.
Ultimate Finish (An expert choice)
(Extra – Problems) + Answer key
(Ultimate Finish) + Hints/Solutions
(a) Ultimate Finish + Extra – Problems Solution + Extra – Problems (Answer key)

SAMPLE PAPERS


(a) SAMPLE Test Paper for IIT- JEE Mains -2015
(b) SAMPLE Test Paper for IIT- JEE Advanced – 2015
Answer key
(a) SAMPLE test paper for IIT JEE - Mains -2015
(b) SAMPLE Test paper for IIT JEE - Advanced – 2015
Solution to sample test paper for IIT- JEE Advanced – 2015
PUZZLE – SECTION - Solution
IQ Problem Solutions
174
184
187
198
204
213
213
213
220
221
vii
viii
Contents
Ramesh Chandra IIT-K
Permutation & Combinations
“The permutations and combinations are endless. It's like a game of three-dimensional chess.”
(Sherry Jeffe)
So first understand the need of topic and the motive of studying like, why are you reading this topic? This will limit
your requirement. Don’t go beyond certain limit until unless you are not researching on topic. The topic is written
in best possible sequence so don’t break it and do not leave any concept, these are the basics building blocks. First
try to solve the entire problem your own and then look for solution. I am sure at the end of chapter you will become
an expert, which lead to make smile at your face. Hope you will enjoy the topic. Best of luck! to your expectation.
1
2
Permutation and Combinations
BASIC REQUIREMENT TO THE TOPIC
FACTORIAL (!)
This is the basic building block of permutation and combination hence knows how to use it.
Factorial n is written as “n!”.
Its meaning is as “the product of first n natural numbers”. i.e.
n! = 1×2×3×……..×(n - 2)×(n - 1)×n. Where n ∈ Natural number
Examples.
1! = 1.
4! = 1×2×3×4=24.
5! = 1×2×3×4x5 = 5×4×3×2×1=120.
8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320
0! Defined as equal to 1 so don’t think why 0! = 1! = 1 just accepts it.
We do not define the factorial of negative integers or proper fractions.
2
3
i.e.  2  !  Not Define ( ND ) ,   ! ND .
Here I am going to ask some True/False questions. Just answer them.
1. Is (2  3)!  2!  3!?
2. Is (5  3)!  5!  3!?
3. Is (5  3)!  5!  3!?
 5  5!
4. Is  !  ?
 3  3!
Here answer to each question is False, Hence from here we get the following results.
Results:
Ex1.
Sol.
1) (m + n)!  m! + n!;
2) (mn)!  (m!) x (n!);
3) (m  n)!  m!  n!;
4) 
 m  m!
.
! 
 n  n!
Find the values of following.
1 2 3
 
3! 4! 5!
I.
10!
2!8!
I)
10! 10  9  8  7  6  5  4  3  2 1 10  9


 45.
2!8! 2 1 8  7  6  5  4  3  2 1
2
II.
III.
5!
5!
5!
5!
5!
5!





0!5! 1!4! 2!3! 3!2! 4!1! 5!0!
Ramesh Chandra IIT-K
II)
3
 1 2
 6  24  120
1 2 3 
    20  10  3
3! 4! 5! 
120


III)
5!
5!
5!
5!
5!
5!





0!5! 1!4! 2!3! 3!2! 4!1! 5!0!
=


 27
9
 .

 120 40


5! 5. 4! 5.4. 3! 5.4. 3! 5. 4!
5!





0! 5! 1! 4! 2! 3!
3! 2! 4!1! 5!0!
= 1 + 5 + 10 + 10 + 5 + 1 = 32. Here we are using the result n! 
 n  1!  n
. Think! Why?
It’s simple to prove.
5! = 5 x 4 x 3 x 2 x 1 = 5(4 x 3 x 2 x 1) = 5 x 4! = 5 x 4 x 3!, and so on.
Now you can say in general n !   n  1  !  n provided n  1  0 .
Similarly n!  n   n  1   n  2  ! provided n  2  0
n
This can also be written as n ! 
k

or recursively defined as n !  
1
if n  0
( n  1)! n if n  0
k 1
 is use for multiplication of all the entities. Let us solve some question to get the theory.
1 1 x
 
6! 7! 8!
Ex2:
Solve for x,
Sol:
8 x
1
1
x
x
1
1
x
8
  So x = 64.
 1  
 


6! 
7  8  7  6!
6! 7  6! 8  7  6!
1 8
 7  8 7
Ex3:
Solve the equation
Sol:
n!
n!

 (3!) 2  1
( n  2)! ( n  1)!

 n.(n  1)  n  1  (3.2.1)2
 n2  2n  1  (3.2.1) 2
n!
n!

 (3!) 2  1 .
(n  2)! (n  1)!
n.(n  1).(n  2)! n.(n  1)!

 1 (3!) 2
 n  2!
 n 1!
 (n  1) 2  (6)2  n  1  6  n  7 or  5 , But n can’t be negative integer so n = 7 is answer.
Ex4:
Find the remainder when1!  2!  3!  .......  n!, where n  10 , is divided by 6.
Sol:
6! 7! 8! .............n !  6I i.e. multiple of 6, so remainder for this quantity = 0.
Now remaining 1! 2! 3! 4! 5!  1  2  6  24  120  6 I1  3 where, I1 is another positive
integer, Hence remainder = 3.
3
4
Permutation and Combinations
SUMMATION
This is the topic, which most student hate at start, I am going to explain it in details so that your fear factor
becomes zero. Hence, first understands the meaning of it, only then we can play with the summation type problems.
5
Ex1:
T
r
in this case lower limit = 1 and upper limit = 5 and Variable function is Tr .
r 1
5
Its meaning is as shown,
T
r
 T1  T2  T3  T4  T5 . (Look, how r is changing.)
r 1
Now let say Tr  (2r  1) then
5
 (2r  1)  (2.1  1)  (2.2  1)  (2.3  1)  (2.4  1)  (2.5  1) .
r 1
 2(1  2  3  4  5)  (1  1  1  1  1) .  2 
1  2  3  4  .....  n 
5 6
 5 = 35. Here we are using the formula
2
n(n  1)
. Let us have another example.
2
10
Ex2:
m
2
 22  32  42  ..................  102.
m 2
Observe the expression, variable m is starting from 2 and end at 10 and variable is increasing by 1 unit
each time.
10
 m  1
2
2
 22  32  .....  102   12 =
m 2

2
2
10 11 21
 1  385  1  384.
6
2
FORMULA RELATED TO SUMMATION
n
1)
2
Here we are using the formula1  2  3  .....  n 
 r  1  2  3  ......  n 
r 1
n(n  1)
.
2
n  (n  1)  (2n  1)
.
6
Ramesh Chandra IIT-K
n
2)
3)
r 1
n  (n  1)  (2n  1)
.
6
n
2
r
2
 12  22  32  .....  n 2 
 n(n  1) 
r 3  13  23  33  .....  n 3  

 .
 2 
r 1
n
4)
 k  nk Where k is any constant.
r 1
n
Ex3:
 2  2  2
 .............. 2  2 n.
 2  2
r 1
n times
n
5)
n
n
n
n
n
  k1r  k 2   k1  r   k 2 or in general
  k1f (r)  k 2g(r)   k1  f (r)  k 2  g(r).
r 1
r 1
r 1
r 1
r 1
r 1
n
Ex4:
  5r  2    5  1  2    5  2  2    5  3  2    5  4  2   .......  5  n  2  .
r 1
 2 
2
 ........ 2.
= 5.1  5.2  5.3  ............5.n  2

ntimes
n
n
r  2.
 2 
2
.......... 2. = 5
= 5(1  2  3  ................  n)  2


r 1
ntimes
= 5
r 1
n(n  1)
 2n
2
n
Note: Observe, how we write
n
n
  5r  2  = 5 r   2 .
r 1
r 1
r 1
Now answer the next question i.e. formula and remember the result.
n
Que1: Show that

n
n
 

  f (r)  g(r)     f (r)     g(r)  . verify it by taking any suitable example.
 r 1
r 1
  r 1

3
Sol.
Let us take an example as
 r(r  1) , here f(r) = r and g(r) = r + 1.
r 1
3
So
3
 r(r  1)  1  2  2  3  3  4  20 &
r 1
3
 r  1  2  3  6,
r 1
 (r  1)  2  3  4  9 ,
r 1
n
Clearly 20 ≠ 6 x 9, Hence, here we can verify that
r 1
100
Ex5:
Find the value of
 r   r! .
r 1
Sol.

n
 
n

  f (r)  g(r)     f (r)     g(r)  .
100
100
100
100
r 1
r 1
r 1
r 1
 r  r !   (r  1  1)  r !     r  1 r ! r !  (r  1)! r !
r 1
r 1
5
6
Permutation and Combinations
2!
3!
4!



1!
2!
3!
5!

4!
...
... = 101! – 1.
= ...
.... ..... ...
.... ...
...
100!  99!
101!  100!


101!1!
Ex6:
Select the greatest Number.
(A) (10!)2.
Sol.
10!
2
(B) (2!)10.
(C) (5!)4.
5
(D) (4!)5.
4
 13168189440000,  4!  7962624,  5!  207360000
Hence correct option is A
Here we have an exercise for you to solve. First complete it and then proceed.
EXERCISE 1
Q1. Find the value of following
a) (6  2)!
b) 6! 2!
6
2
and verify that (6  2)!  6!  2! . Is   ! 
6!
? What do you learn from this question?
2!
Q2. Convert the following into factorials:
a) 6 x 7 x 8 x 9 x 10
Q3. Evaluate:
b) 3 x 5 x 7 x 12 x 32
n!
for the following
r !( n  r )!
a) n = 10, r = 2.
Q4. Prove that
b)
n = 5, r= 3.
(2n)! n
 2  (2n  1)  (2n  3)  (2n  5)  ......  5  31. Hence deduce that
n!
30! 15
 2  29  27  25  ......  5  3 1.
15!
ANSWER KEY
1.(a) 24 (b) 718
2. (a)
10!
(b) 8!
5!
3. (a) 45 (b) 10
Ramesh Chandra IIT-K
BINOMIAL COEFICIENTS ( n C r ) AND SOME IMPORTANT FORMULAES
Here, n C r 
Ex1.
n!
, where n and r  Whole numbers such that 0  r  n .
r!(n  r)!
Find the value of following (1 - 4)
1)
10
5
2)
C2
C3
3)
7
C5
4)
n
C1
5)
Prove that 13 C 3  13C 4  14 C 4 , and hence deduce the Pascal identity n Cr  n Cr 1  n 1C r 1 .
6)
Prove that C5 
14
n
Sol.
Cr 
1)
10
14 13
14 13
14 13 12 11 10
 C4    12C3     
and hence deduce that
5
5 4
5 4 3 2 1
n n 1
 Cr  1
r
C2 
10!
10! 10  9  8!


 45.
2!(10  2)! 2! 8!
2! 8!
5!
5!
5  4  3!


 10.
3!(5  3)! 3! 2! 2! 3!
7!
7!
7  6  5!
7
3. C5 


 21.
5!(7  5)! 5! 2!
5!  2!
2.
5
C3 
4.
n
C1 
5.
13
n   n  1 !
n!

 n.
1!(n  1)!
 n  1!
C 3  286, 13 C 4  715, 14 C 4  1001. Clearly 286 + 715 = 1001.
n!
n!
Now n C r  n C r 1 

r!(n  r)! (r  1)!(n  r  1)!
n!
1 
 1

=


r!(n  r  1)!  n  r r  1 
n!(n  1)
(n  1)!
=
=
= n 1 C r 1 .
r!(n  r  1)!(n  r)(r  1) (r  1)!(n  r)!
14! 14 13! 14 13! 14 13

 
  C4
5!9! 5  4! 9! 5 4! 9! 5
13 12
14 13 12
14 13 12 11 10
13
14
Similarly C 4   C3 so C5 
  C3 =      9C0 . Here 9 C 0  1
4
5 4
5 4 3 2 1
n n 1
n!
n  (n  1)!
n
(n  1)!
Now n C r 

 
=  C r 1 .
r!(n  r)! r  (r  1)!(n  r)! r (r  1)!(n  r)! r
 n   n  1   n  2  n 3
n
Note: This property can be extended more as Cr     

 Cr 3 , provided n ≥ 3, r ≥ 3.
 r   r 1   r  2 
6.
14
C5 
Now we have a question at the moment, if i say that
(i)
7! 7
 C2
2!5!
(ii)
7! 7
 C5
2!5!
(iii)
7! 7
 C2 Then which one is correct?
2!6!
7
8
Permutation and Combinations
Here first two options are correct. Observe here 2 + 5 = 7 but 2 + 6 ≠ 7.
From here we leads to the following results
(x  y)! (x  y)

Cx  (x  y) C y
x!y!
OTHER FORMULAES
n
C x  n C y Then, this implies either (i) x = y OR (ii) x + y = n.

If

n
C r  n C n  r . {since x = r, y = n – r, x + y = n}

n
C r  0 , if r > n. It is taken by convention.

n
C0  1,n C n  1

n
C0  n C1  n C2  n C3  ..........  n C n  2 n .

(1  x) n  

n  r 1
Cr x r so coefficient of xr in 1  x 
n
is n  r 1 Cr .
r 0
Observe that all the power of x  {0, 1, 2, 3,……..,∞} i.e. whole numbers.
m n  l

(1 + 2 + 3 + 4 + ………. + n)2
2
Cm 
2
n l
Cn  lCl 
(m  n  l)!
.
m!n!l!

2
2
= 1  2  3  .........  n  2(1.2  1.3  ...1.n  2.3  2.4  .......  (n  1).n)

A geometric progression a, ar, ar2, ar3……… ar(n-1) . Sum of its first n terms
i.e. Sn = a + ar +
ar2
+
ar3 +…+ar(n-1)
 1 rn 
= a
.
 1 r 
If |r|<1 then S∞ = a + ar + ar2 +……….+ ∞ =
a
.
1 r
Have you memories the above formulas? Let us have a mind refreshing IQ problem.
IQ 1:
An ant wants to go from A to B by travelling on the surface of cube of
length 1 cm as shown in figure. Find the shortest distance travel by ant.
Note: See solutions at the end
SOLVED EXAMPLES TO EXPLORE ABOVE FORMULAS
Ex 1:
If n C 3  n C 7 then find the value of n.
Sol.
n = 3 + 7 = 10. {Since n C x  n C y Then this implies, either (i) x = y OR (ii) x + y = n.}
Ex 2:
If C9 
Sol.
by property Cr     
n
12 1110 9
 C6 then find the value of n.
98 7
n
 n   n  1   n  2  n 3

 Cr 3
 r   r 1   r  2 
10 9
11
12
 C6  10C7 ,  10C7  11C8 ,  11C8  12C9 
7
8
9
n
C9  12C9  we can say that n = 12.
Ramesh Chandra IIT-K
Ex 3:
Find the value of n C 0  n 1C1  n  2 C 2  n  3C 3  n  4 C 4  ..........  2 n C n .
Sol.
Apply n Cr  n Cr 1  n 1C r 1
We can write n C0  n 1C0  n 1C0  n 1C1  n  2C1 and
At the end we will get
2n
n2
C1  n  2 C2  n  3C2 and so on.
Cn 1  2 n C n  2 n 1C n .
So n C0  n 1 C1  n  2 C 2  n  3 C3  n  4 C4  n  5 C5  .........2 n C n  2 n 1C n
Ex 4:
Prove the Newton’s identity n C i × i C j = n C j ×
Proof:
n
Ci  i C j 

n- j
C i- j for integers 0  j  i  n .
n !i !
n!
( n  j )!


i !( n  i )! j !(i  j )! ( n  i)! j !(i  j )! ( n  j )!
n!
( n  j )!
n- j
= n C j × C i- j .

j !( n  j )! (i  j )!( n  i )!
It would be good if you remember this identity.
NOTES MAKING
HOW TO MAKE NOTES?
When you read the book, you will have some thinking and questions in your mind. There might be
situation that you want to discuss to your teachers. Even you can have a better solution to particular
question, and then write it for your future. You can discuss your thinking at [email protected]
Que No./
Page No.
NOTES
9
10
Permutation and Combinations
POINTS TO REMEMBER:
 .
 .
 .
 .
 .
 .
 .
 .
OTHERS: ----------------------------------------------------------------------------------------------------------------
Contents
PERMUTATION AND COMBINATIONS
Book for Boards, NTSE, IIT JEE- Mains/Advanced, CAT, Olympiad & Software Engineers
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