Permutation and Combination _ Class Notes __ Manzil Legends-JEE.pdf

P n c notes for jee main

1. MANZIL

TARUN KHANDELWAL Sir

Permutation &
Combination

2. TOPICS TO BE COVERED

[1 } Fundamental Principle of Counting
a Permutation of Distinct Objects 4 Alike Objects

13.) Combinations

[4] Selection of At-least one object

[5. Groups, Inclusion Exclusion Principle 4 Beggar's Method

BB oe-arrangement

Arras 2021 a 2022

3. LC

>

V If the total number of ways of occurrence of event A

Fundamental Principle of Counting 6

and the total number of ways of occurrence of another event B is "n”
ways then the total number of ways of simultaneous occurrence of

both the events is "m.n a Shi Ti ,Sh D
SEE sur | shat.

a Teams 0 vine | Shak.

5 I ways = Bx2.

If the total number of ways of occurrence of m A is "m” ways,
the total number of ways of occurrence of another event B is 'n"
ways & the total number of ways of occurrence of another event C
is "p" ways then the total number of ways of simultaneous ati
occurrence of 3 events is "m.n.p”

4. are not allowed to repeat in any number formed by using the digits
en the number of all numbers greater than|10,000 Is equal to.

[2021 Main, 22 July 1]

[Ans. 96]
EL
4 4

Tell rumbu, =

HxUxaxax)

ur

a= À

PN
sy

5. OY
Find the number of @)digit integers divisible by|3|which can be formed using ®

the | neh 2 0% 2 3 4, an without repetition.

u FEB];

Ss a Bs 4d
De = sl=120
‚u: on
HU Sr
Tati = 4
hm 204% we = 16

TAC
—

6. aaaa d

The total number of 5-digit numbers] formed by using the digits

without repetition, which are multiple ofGjis

36 Ame

48

60 Ce ic ts

x

Sum
[2022 Main, 28 June I]

Geol ais ASS
2

1,2.3,5,6,7

[Ans. D]
(EE
ui x <= 4

: Kun:
==

7. Let S be the set of all passwords which are six to eight characters long, where
each character is either an alphabet from (4, B, C, D, E) or a number from (1, 2,
3, 4, 5} with the repetition of characters allowed. If the number of passwords
indy whose at least one character i is a number from (1, 2, 3, 4, 5) is@ x 58) then
ais ae to A

TR ET

- [Ans. 7073]
& (1° 409] SAS AREA
AA — Ñ
te 1elorin}— 8 (1+ 55" 2
sl) E SN (a
(pers
S (ua) a <= (ét 30)

of

)

®

8. Find the number of 4 digit integers in the closed interval [2022, 4482] which
o can be formed using the digits (0,2,3,4,6,7). e
TEE Advanced 2022]
LE A — +: ANS 569
= 569

Acuna Lia
ERED)+e GLL|)e

9. 5

Aw) uw)
Numbers are to be formed between 1000 and 3000, which are divisible by 4, ®

using the digits 1, 2, 3, 4, 5 and 6 without repetition of digits. Then the total
number of such numbersis___.

[2022 Main, 26 July 11]
[Ans. 30]

10. a
lll Theorems of Permutations for Distinct Objects
v

Theorem 1

Theorem 2

The total number of permutations of n distinct objects taken all at a
time with repetition is = n”

ER we
The total number of permutations of n distinct ‘objects taken all ata

time without repetition is = n!
A ees

The total number of permutations OFF distinct att taken "p" ata

= nf ass
time with repetition is = n Hs

The total numbers of permutations of r i Ge. out of n distinct
objects without repetition i:

This number is denoted as

11. Find the number of ways in which 5 boys and 5 girls can be arranged in a
line such that-

(a) All the boys are not together
(b) No 2 boys are together.
(c) Boys and girls are alternate

12. the letters of the word 'VOWELS', so that all the consonants never come
=

| |

Th f six | ds (with ithout ing), fi d using all
lo, e number of six letter words (with or without meaning), formed using al ®

together, is.

) [Ans. 576]

él - 314) LL

> El - 6x4 ELS
= ul [s-1) mh

x
= Gaal ny =

24xaU = SFE
um

+E

13. Ss. A
Ci or te je ¿50

Theorem 5

(Permutations of objects not all distinct) Sl

31 4.21
Let we have 3 identical A's and 4 identical B's and 6 identical C's ei

On
!

13!
then total number of arrangements of these 13 alphabets is ——. al

— st
2121
In general, Let there be n,> A,'s, nz > A2'S,n3 > Ag'S....., Np > Ag'S

(M+n2+..+ng)!

Then the total number of permutations =
Ny! Ng! ... Nk!

14. only is

The number of Loa ac integers with sum of the digits equal to@Qand

formed by using

> = “12021 Main, 26 February 1]

E)

= 42 {ans A]
as Cu
35 nn ~

82 nes

15. )

The number of 5-digit natural numbers, such that the product of their digits is ®
36,is__

[Ans. 180]

16. lo) The total number of three-digit numbers, with one digit repeated exactly two
times, is

Ca ee
=: [Ans. 243]

17. a
{e | Rank According to Dictionary

If all the letters of the word ‘VINOD’ are arranged in all possible manner as
they are in a dictionary, then find the rank of the word ‘VINOD’.

; ; ID I Nov

NCS
à - D
= Inwvo

VET) = 31 V+ orar
VID = loch werd.
D- 2: m
VINDo… , ;
VUNoD —ı

<

18. y
If all the letters of the word “MUSTAFA are arranged in all possible manner ®
as _— areina i then find the rank of the word en

eT En USA DORA, En satu
SE
Frans — CRU st [ED say .

MA OT) > 5/=120 MUSTA RF ]=¡ - |
ME TE) > Sat +66 MUSTAFA —4( =;
MS EJ Saz Guante
o ss

19. )

arranged in serial order as in an English dictionary. Then the serial number of
w) the word'MANKIND'is _.
(u ie

the letters of the word 'MANKIND' are written in all possible orders and ®

[2022 Main, 25 July I]
[Ans. 1492]

20. If N= 114 214 31+41+ (Qi) zo 4
then N cannot be a perfect square for x> 4. 371
d 6
ES ía O7 eas
+20 = % Bo 6
| 7-9
ae OX E
| FR ar
Me eh Shel Le
9 +03) q

O es Ta a 6
334 Si

21. fl @
El Properties of Combinations [EVA] =
E

v "Le!

1. "Cy="C,=1
Vol
CELL Ex» PC + Ge Ce
; =
nn) ET ee. 13
4. nc, =n Les A a

5. ™P,="C, «rl

++ 6. tp 4

22. dddd I

y
A scientific committee is to formed from 6 Indians and 8 foreigners, which
includes at least 2 Indians and double the number of foreigners as Indians.
Then the number of ways, the committee can be formed is:

CEE RYE) or EUER) oy („1202 Mao 24 February

560 pS 4Z& GE) [Ans. C]
6 8 € 3 6
1050 Ct GR ARA Sn
OR — (+)
1625
575

)

®

23. yo Y

number of ways of forming a group consisting of three boys and three girls, if

lo, There are ten boys, B,, B,, ..... Bj) and five girls G,, G,, .... G; in a class. Then the
both B, and B, pa should not be the members of a group, is

Total — when (B, hu on aie ne = re
5
("cx G)— BC a Seg
Sr, - sa
= | Lox 9x8
E ( Rz - 8]

lo [ 12 8- 8]

lox iz
= 1120

24. There are 15 players in a cricket team, out of which 6 are bowlers, 7 are Ly
o batsmen and 2 are wicketkeepers. The number of ways, a team of 11 players

be selected from them so as to include at least 4 bowlers, 5 batsmen and 1
E 7

me Murs) is a Y

eee ne...

ve MES Bol ys Bok K1W) oF 25% € Bat kin) oa 12021 Main, 20 July 1
at (s Bat & zu) 2771
+ G RICE +(éc a (UBKS

a 1 ) Y. (CA Fo

Gx 21%) + (€ ISX 4x2) Ec, REAL
ad a2 Clsxzıxı)
al [a + lot 15) woe

Ql x 34| = $77

25. lok
”
®
®

If the sides AB, BC and CA of a triangle ABC have 3,5 and 6 interior points
pectively, then the total number of triangles that can be constructed using
(esapoins as vertices, is EN to:

CA ETS EN T2021 Main, 17 March I]
Ca i Ce + °Cn + 3) [Ans.3]
364

240
333

360

O N

26. which exactly one is correct. There are 3 marks for each correct answer, -2
marks for each wrong answer and 0 mark if the question is not attempted.
Then, the number of ways a student a in the examination ge

lo, In an examination, there are@)nultiple choice questions with 3 th 3 choices, out of ®

marks, is . G3, De
. our 712022 Main, 24 June 1]
[Ans. 40]
De a an
Ss x. @). (e. I
3 nd A)
= loa ga ug -8)

= +0

O N

27. )

and 2 girls from the class is 168, then b + 3g is equal to A
(Hv)

D A class contains b boys and g girls. If the number of ways of selecting 3 boys ®

“fans. 17]

28. Team 'A' consists of 7 boys and n girls and Team 'B' has 4 boys and 6 girls. If a
total of 52 single matches can be arranged between these two teams when a
boy plays against a nik and a a FL against a = then n is equal to

RUZ

“2021 Main, 17 March m
[Ans. C]

a

>

a

2094 ad

Y
sy

29. il Important Note EC, = no of Sido + no 4@
{|| nn Ni c= "A SMS + roof dun

“NOTE The total number of diagonals in a “n” sided Polygon is

“a

“Nore Total number of subsets of a set containing “n” distinct elements is

30. La]

él

ex [Tool] cout 5
Ex pomant = (+ cmt ales + (35 Eo

cta”

x= 24,
Sigena sf ee in lol
Pe | E (12). +]
tt
¡di

&

31. Find the number of zeros at the end of 100!.
—_“_

Jo — axs

a 24
loo! = @ . G)
\ _5 sl =i2©

Ae = 2 (eo) e ood

= 5 2

201, => Go)” > 822

32. 2 ®
El Formation of Groups MIS objects) dd

v m No à 3
Wen) Se var
SY nm mint |
Min 5 5,
rn nl l
Qn) > n un + EI, 7 “
\ 1€ lolle/GT)
Lol

E ==
EN
ne QG | N

33. Gil Formation of Groups- Theorems
v

Theorem 1

The number of ways in which (m + ») different things can be divided

into two groups such that one of them contains m things and other than
Ñ . (m+n)!

has n things, is Sainte (mn).

If these groups are the be “distributed” between two persons then the

number of ways= za) 21)

min!

The number of ways in which (7) different things can be divided into

2n)!
two groups such that EACH of them contains n things is am

n!n!

If these groups are the be “distributed” between two persons then the
(2n)!
number of ways = en

nin!(2!) S

34. = 201
EY sal AD
lal. %
i a
Yi á .
El zul

ees: “ere
DE.
CERN

&

35. Gil Formation of Groups- Theorems
v

Theorem 3

Number of ways in which (m + n + p) different things can be divided
into three groups containing m, n & p things respectively is:

(m+n+p)!
m!n!p! id MARA

If these groups are the be “distributed” between three persons then the

_ (m+ntp)!
number of ways = ap x3!)
Y

nl __ Gn)!
If m= n= p then the number of groups = amar oO

ES
If these groups are the be “distributed” between

Gn)!
number of ways = PO) x3!)
n!n!n!

ree persons then the

36. In how many ways can 6 distinct balls be distributed to 4 boxes such that no
box is empty.

hm

SR

Uno =F LÉ:
= Boxle. 37 ‘
ai

Gh x al ei se @)

G Te | 2 = x eg

O0 >
- = Fora
1630436033 TS

Us] = 3603

Be
B3
Bu

)

®

37. a @
Selection of At least one object er
{|| At but à = al ="
Case 1 : From(n Distinct objects

-

38. >

| Selection of At least one object

&
há Case 1 : From n Distinct objects (E 1)

Total number of ways of dealing with n distinct objects is (2").

39. From a library containing 4 different physics books, 5 different chemistry @
books & 6 different maths books. Find the number of ways in which we can

select ie Pi, Po, Pe, Py, © Ca G--Ce, My, Ma.
(a) Atleast 1 book — œ 1) O &
——

(b) Atleast 1 book of each Subject QL DA Ge 1) x (E 1) .
Is is
(© Atleast 2 books (a -1-1)=@ — 16)
(d) Atleast 2 book of each Subject Q4_\- 4) y(2%1- -S)y QE 6)

(e) Exactly 2books IS Gs). ak ey (att)
A | %

(f) Exactly 2 books of each Subject ASEO
he ae nee ean eue cesses nennen een) Tin nee sencsescecenseeceesesceesucececeseences

40. Total number of ways of dealing with n identical objects is (n+1).
no

41. yo Y
From a library containing 4 identical physics books, 5 identical chemistry ®
& books & 6 identical maths books. (eppHcce of ways in which we can

select hcccccMMMMMM
(a) Atleast 1 book (6x6x+ =) ee mr an eo =

(b) Atleast 1 book of each Subject
= CCE (I) = Hxsxe
Atleast 2 book.
(c) Atleast 2 books rd Be Geta)

(d) Atleast 2 book of each Subject E
a E

= Exactly 2 books
>

=

(f) Exactly 2 books of each Subject

42. books & 6 identical maths books. Find the humber of ways in which we can

select =o,
(a) Atleast 1 book

OY
lo, From a library containing 4 different physics books, 5 different chemistry ®

fi Pa Pa Py CG Cra Cute MMM. M

(b) Atleast 1 book of each Subject
(c) Atleast 2 books
(d) Atleast 2 book of each Subject
(e) Exactly 2 books

(f) Exactly 2 books of each Subject

43. il o
PR e
ll Finding Number of Divisors =
Vv

q = IS
oe
NE +} T6o0/ - Boe a ESE, u IL
Fan Ps? ° |
£ 2a
yA ae
“S ES

TR no divisors CCS = laa.

Lege |

44. In the previous example, out of these 120 divisors of the number N = 75600,
a) How many divisors are even.
@ x 2% = Lx 4xrx2r

(b) How many divisors are odd E .

c) How many are proper divisors = 1&o- 4 =
(c) How many are proper divi 20-2 =D

45. a

© Io Io Jo.

&

A natural number has prime factorization given by n = 2* 3Y 52, where y and z
—o

5
are such that y +Z=5 and y! +z"! ==, y > z. Then the number of odd divisors
E 6 _——
of n, including 1, is:

[2021 Main, 26 February II]

u 9 Ne ES

6x % La 2°

5 5 ba , dd divisors = wu
yz = A = 12

46. =)
&

El Distribution of Identical Objects (Beggar's Method)

In how many ways can we distribute 8 identical coins to 5 beggars such that
(a) each beggar can receive any number of coins. ne.

<

rar

0000400008 Reo Fake -
G, o, rar o) Totes ne ey 0 Le
alu
E LEE CP
AS

79

47. a
lll Distribution of Identical Objects (Beggar's Method)
v

Method
STEP-1 Introduce fake identical coins, 1 less than the number of beggars.

STEP-2 Arrange these fake coins along with the real coins to get the number
of ways in which the distribution can be done.

&

48. a
{e I Distribution of Identical Objects (Beggar's Method)
Vv

¡UTA Each arrangement of these (fake coins + real coins) corresponds to
1 distribution pattern. zZ in

AUIWA The fake coins act as partition between the number of coins received

by each beggar.

&

49. La]

{e | Distribution of Identical Objects (Beggar's Method)

la In how many ways can we distribute(10) identical coins to(S)beggars such

that each beggar receives at least one coin.

<

Aw = a = Te |

&

50. Gil Theorems- Beggar's Method
Vv

Theorem 1| Number of ways to distribute “n” identical objects amongGr)persons so

that each may get any number of things is equal to "*""1C,_,..-

Ts =>
Kar

Theorem 2| Number of ways to distribute('n)identical objects among “r” persons so

that each gets at least 1 thing is equal to A
me Die } NL
+l
) — Fake ha E Yu

¥-1

51. >

{e Finding Number of Non Negative Integral Solutions

V2 nan-v&
Find the number of Integral solutions for the following:
1. x, +X, +X,4+X,+X > atte _ a
tee OT ? =.

2. X, +X) +X, = 50 re ar

2 2.

52. a
{e | Finding Number of Integral Solutions

Find the number of Integral solutighs fo Gale
a.) 3: x +X) +X, +X, 4206 > NE > D (622%, 24)
2 Ss E seo 4 ee 21¢

Xs>0
4. ee 3 x
= Y Y wa

1343 Cy =], |e

te See hs 218.

x
LAM EX 18

4 0.

5: %+%+%,220,%20

6. X,+X,+X3+3x,=20,x,>0

53. In a unique Cricket series between India & Pakistan, they decide to play on
till a team wins 5 matches. The number of ways in which the series can be
won by India, if no match ends in a draw is:

—

© © © O®
Be: EN

Xi we

O or Kat Xa koa Xe Y

Sy 4,
(= Ge Ara,

54. can be placed in a row so that between any two red cubes there should be at

OY
lo, The number of ways, 16 identical cubes, of which 11 are blue and rest are red, Q
least 2 blue cubes, is

ae à Ral) 12022 Main, 27 june 1

[Ans. ze
RL aL al
ED Ke
ality = Xue SEX a
— 5
xro NEM, Ks YE
X or

3
Beate Wie
y

= se"

55. The total number of positive integral solutions (x, y, z) such that xyz = 24 is

[2021 Main, 25 February I]
[Ans. D]

Room

lox3= 30

nN ca
+ a

w
So

ddda
7)
3

&

56. “a
{ | I Circular Permutation (of distinct objects)
v
» The total number of circular arrangements of n distinct objects is =

LE

« If clockwise and anti-clockwise arrangements are considered to be same then
the number of arrangements is

57. a
{|| De-arrangement/Mismatching ie

PE

Libre ta Pise. OU Ea) elle)?
Ej E Ene, En

A

>, E LE Lei}

Leis a

&

58. E)

E Ly Lata De 7
E Er Es Ey

— (ucuoW) + Gouin) + Go Kr) + (cam) +
al = (uexon € ) e
:
ALO + Où GX Ds + up, + ©
le ( u i =;
o Ww 4 Min ud 5
a =

V+ 64 8+>x

o Ms MY-is=9
Sintloody Diet Re

=

59. nr
CR
v

+ A de-arrangement of 1, 2, ..., n is a permutation of the numbers such that no
number occupies its natural position.

ES 1 1 1 (-1)"
. D, =n! [1-F+h-=G+ + |

A 3! n!

60. )

their bags outside the exam hall, while their way back home, in how many
ways can they pick up one bag each randomly such that-
(i) No student ends up picking their own bag. => D¿ Y

(ii) Exactly 2 students pick up their correct bags. — €c, xDy = 1Sx 32135

(iii) At least two of them pick up wrong bags.
a ee

lo, 6 students who are appearing for IITJEE Exam in a centre are asked to leave ®

im) M-I ES ELLES qu &ıc suktc bY
Better oppreach cl o
; = (RBk6C + Iwuse
= [obres + 1usc)

y

Has Cdi 6:
= FQ Ju

61. Thank You !

# IIT Phodna Hai

&