1. How to count without actual counting
Elegant technique of counting
(1) General introduction & Historical development : Useful in trade, business,
industries, Govt. sectors and also the study of combinatorial coefficients.
(Coupled with a story) A grandfather had 6 school going grand children ...........
(2) Story leading to fundamental principle of counting : If an event can occur in m
different ways following which another event could occur in n different ways, then
total number of ways of simultaneous occurence of both the events in a definite order
is m × n (This can be extended to any number of events)
Models: Number of ways in which one can travel from town T1
to T3
(a) via Town T
T2
in a definite order = 3×2.
(b) Cinema Hall
(i) Number of ways in which he can enter and leave the cinema hall by a different
door = 5×4
(ii) He can enter & leave by any door = 5×5 = 25.
(iii) He can enter by D1
, D2
and leaves by D3
, D4
, D5
= 2 × 3 = 6
(c) Tossing of a coin & Tree diagram
(2×2×2)
(d) Keeping P | C | M | E book on a shelf. = 4 × 3 × 2 × 1
(3) Statement of Fundamental principle of counting supported by graded examples.
Examples on Fundamental principle:
(i) 3 digit numbers using the digits 1, 2, 3, 4, 5 without repetition. (5.4.3 = 60)
(ii) 10 T/F questions. How many sequences of answers are possible. (210
)
(iii) 10 students compete in a swimming race. In how many ways can they occupy the first
3 positions. [ Ans : 10. 9 . 8 = 720]
(iv) 7 flags of different colour. Number of different signals that can be transmitted by the
use 2 flags one above the other. [ Ans: 7. 6 = 42]
A well known nurserypoem justifying FPC says
– I met a gentleman with 7 wives
– Each wife had 7 sacs
– Each sack had 7 cats
– Each cat had 7 kittens
– How many kittens did the traveller meet.
[ Ans : 74
= 2401=]

2. Machine/Miracle if Vowels may occupy the
(a) odd position. (4 · 3 · 2) (4 · 3 · 2 · 1)
(b) even position. (3 · 2 · 1) (4 · 3 · 2 · 1)
(vii) 4 lettered word using only the letters from the word “ DAUGHTER” if each word is to
include “G”. (4×7.6.5 or 8.7.6.5 – 7.6.5.4)
(viii) A letter tock:
(4) Significance / meaning of the title of the chapter.
Permutation & Combination and introducing notion of factorial.
Permutation means arrangement in a definite order of things which may be alike or
different taken some or all at a time. Hence permutation refers to the situation where
order of occurence of the events is important.
DEFINE : Things which are alike and which are different. All God made things in
general are treated to be different and all man made things are to be spelled whether like
or different.
Combination – on the other hand refers to the situation where order of occurence of
the events is not important.
0! = 1 ; 1! = 1 ; 2! = 2 ; 3 ! = 6 etc. (factorials upto 7 must be remembered)
(2n!) = 2n
.n! [ 1.3.5........(2n–1)].
Note that sum of all the factorials > 4 is never a perfect square.
Examples: (1) If (100)! = 2m
.I where I is an odd integer then find m
(2) Find the number of cyphers at the end of 100!
(5) Useful theorems( For faster execution rate of the problems)
T–1 : Number of permutations of n distinct things taken all at a time symbolysed as:
n
Pn
= P(n,n) = n
n
A = n!
T–2 : Number of permutations of n distinct things taken r at a time 0  r  n as :
n
Pr
= P(n,r) =
)!
r
n
(
!
n
An
r

 ; note that nP0
= 1 ; n
Pn
= n !
T–3 : Number of combinations/selections of n distinct things taken r (0 < r <n) at a time =
n
Cr
= C(n,r) = 





r
n
= )!
r
n
(
!
r
!
n

n
Cr
+ n
Cr–1
= n+1
Cr
Note that : (1) n
Cr
= n
Cn–r
; n
C0
= 1 ; n
Cn
= 1
(2) n
Cx
= n
Cy  x = y or x + y = n (In a plane there are n circles...)
(3) n
Pr
= r! n
Cr
(4) In a birthday party 101 people. Number of handshakes. [5050]
Example on n
Cr
& n
Pr
(i) There are n points in a plane, no 3 of which are collinear. Find
(a) Number of straight lines by joining these points. (Ans: n
C2
)
(b) Number of triangles (Ans: n
C3
)
(c) Number of diagonals in a polygon = n
C2
– n =
2
)
3
n
(
n 
(d) 10 points in a plane no 3 being collinear except 4 which are in the same line.
Find (1) no. of lines ; (2) no. of triangles
[Sol. (i) 10
C2
– 4
C2
+ 1 or 6
C2
+ 6
C1
. 4
C1
+ 1
(2) 10
C3
– 4
C3
or 6
C3
+ 6
C2
. 4
C1
+ 6
C1
.4
C2
]

3. [ Ans : (i) 7
C2
– 3. 3
C2
+ 3 = 12 + 3 = 15 = 6
C2
; (ii) 7
C3
– 3 = 32 ]
Note : 5 – Pentagon, 6 – Heptagon, 7 – Heptagon, ...... 12 – Dodecogon, 15 – quindecagon
(i) 6 | | lines which are cut by another set of 5 | | lines. No. of parallelograms.
(Ans: 6
C2
. 5
C2
=150)
(ii) In how many ways can the seven different colours of a rainbow be arranged so that the
blue and green never come together.
(iii) A Grandfather with 7 grandchildren takes them to zoological garden without taking the
same three children together. How frequentlyeach child goes and how frequentlyGrand
father goes. [Ans: 7
C3
; 6
C2
]
(iv) Number of words with 10 different consonants & 4 different vowels if each word has
3C and 2V and begins with a vowel.
(10
C3
.4
C2
). 3
C1
.4! or 10
C1
.9
C2
.4
C2
.4!
(v) In a morse telegraphy there are 4 arms & each arm is
capable of taking 5 distinct positions including the
position of rest. How many different signals.
(A) 17 (B) 14 (C) 11 (D) 20
(vi) Boat problem : (General nomenclature). An 8 oared boat to be manned from a crew
of eleven of whom 3 can only steer but can not row. 8 can row but can not steer. In
how many ways the staff can be arranged if 2 of the men can only row on bow side.
1
2
8
7
6
5
3
4
[Sol. – position for steering can be selected in 3
C1
ways
– Two positions on the bow side can be selected in 4
C2
ways and AB can be
arranged in 2! ways .
– Remaining 6 people on 6 positions can be seated in 6! ways
Hence using FPC Total ways 3
C1
. 4
C2
.2! . 6! = 25920 Ans]
[Alternatively: 2 out of 6 people can be selected in 6
C2
ways. Now (6
C2
+ AB)
i.e. 4 persons on bow side can be seated in 4! ways .
Total ways = 3
C1
. 6
C2
. 4! . 4! = 25920 Ans ]
(vii) 4 B and 4 G are to be seated in a line find
(a) Number of ways they can be seated so that ‘No two girls are together’.
(or girls are separated); (b) not all the girls are together or at least one girl is separated
from the rest of the girls; (c)‘boys and girls are alternate’ or boys as well as girls are
separated or each B follows the G and each G follows the boy ; (d) B and G are alternate
and a particular B & G are never adjacent to each other in anyarrangement.(e) If they are
4 married couples then the number of ways if each couple is together. [(2!)4
. 4!]
(viii) Numbers of ways in which 5 different books can be distributed between R/S/G if each
child gets at least one book. [35
– [3
C1
(25
– 2) + 3
C2
]
( note that 5
C3
. 3! . 32
is wrong think why ? )

4. class (K classes). How many different sets of tickets they may have had. [ Number of
different tickets of one class = 10
C2
= 45
 Number of different tickets of k classes = 45k  45k
C6
]
(x) A train having 12 stations enroute has to be stopped at 4 stations. Number of ways it
can be stopped if no two of the stopping stations are consecutive.
[Sol. S1
S2
S3
........ S12
(8 remaining with 9 gaps) select any 4 out of 9 i.e. 9
C4
and this is the final answer.
think! how?
(xi) 8 straight lines & 6 circles in a plane. Maximum number of their intersection points.
(xii) A cricket team comprising of A,B,C,..........J, K is to be sent for batting. If ‘A’ wants
to bat before ’J’ and ‘J’ wants to bat after ‘G’. Then the number of batting orders if
other players could go in any order.
J
J
A
G
G
A



2 ways
[ X X X .... X (11 croses). Select 3 places for A G J who can be seated in two ways and
remaining 8 in 8! ways. Hence total = 11
C3
. 2 . 8! .]
(6) Formation of groups (distribution ofdifferent things among persons are veryconvinient
by forming the groups)
(a) m + n different things m n ; number of groups = !
n
!
m
)!
n
m
( 
If m = n then number of groups = !
2
!
n
!
n
!
n
2
If these groups are to be distributed between two persons the number of ways = !
2
!
2
!
n
!
n
)!
n
2
(
(b) |||ly
(m + n + p) different things can be divided into 3 unequal groups is !
p
!
n
!
m
)!
p
n
m
( 

(1) If the groups are all equal then the number of way =
!
3
)
!
n
(
)!
n
3
(
3
(2) If these groups are to be distributed between 3 persons then !
3
)
!
n
(
!
3
.
)!
n
3
(
3
Proofs and explanation of above
(a) To find the number of ways in which (m + n) different things can be divided into two
unequal groups, it is equivalent to select 'm' persons. Since for each selection of 'm'
persons there will be a corresponding rejection of n persons hence each selection of m
and a corresponding rejection of n people will give a group.
 Number of groups = m+n
Cm
= !
n
!
m
)!
n
m
( 
. However situation is different if m = n.
consider 4 different toys T1
T2
T3
T4
When T1
T2
is selected and T3
T4
is rejected  one way of forming the group.
When T3
T4
is selected and T1
T2
is rejected is not a different group hence !
2
!
2
!
4
gives

5. = !
2
!
n
!
n
!
)
n
2
(
...(1)
However if these 2n things are to be distributed between two persons then (1) has to be
multiplied by2! again.
(b) To understand the article consider 10 children to be divided into three unequal groups of
2,3 and 5.
(i) First make two groups of 2 and 8 and this can be done in !
8
!
2
!
10
way say. AB/CDEFGHIJ
Consider our such group of 8 which can be divided into two groups of 3 and 5 in
!
5
!
3
!
8
ways. Hence total = !
8
!
2
!
10
. !
5
!
3
!
8
= !
5
!
3
!
2
!
10
Similar explanation will be valid if initial groups in 3 and 7 and then split 7 in 2 and 5.
However if 10 is divided into two groups of 5 each initially, which can be done
in !
2
!
5
!
5
!
10
ways ....(1)
One such grouping is say
A B C D E F G H I J
Consider F G H I J keeping AB C D E as it is. Now the group F G H I J can be divided
into two groups of 2 and 3 is !
3
!
2
!
5
and similarly when F G H I J is kept as it is,
A B C D E can be divided into two groups of 2 and 3 in !
3
!
2
!
5
ways. Hence one group
(each of 5) given by (1) generated 2. !
3
!
2
!
5
different groups of 2, 3, 5.
 Total no. of groups =
!
2
!
5
!
5
!
10
.
!
3
!
2
!
5
.
2
= !
5
!
3
!
2
!
10
Similarly if m = n = p situation becomes different.
Consider T1
T2
T3
T4
T5
T6
to be divided into 3 equal groups.
When we say 6
C2
. 4
C2
= !
2
!
2
!
2
!
6
is totally wrong why?
Selected in 6
C2
Selected in 4
C2
Rejected in 4
C2
(A)
T1
T2
T3
T4
T5
T6
T1
T2
T5
T6
T3
T4
T3
T4
T1
T2
T5
T6
T3
T4
T5
T6
T1
T2
T5
T6
T1
T2
T3
T4
T5
T6
T3
T4
T1
T2
R S G
Note that all these six groups are counted in (A), however they are identical. Hence the
answer in (A) is as many numbers of times more more as many number of times these
equal groups can be arranged i.e. 3! times. Hence the correct number of groups is
equal to !
3
!
2
!
2
!
2
!
6
.

6. be = !
3
!
2
!
2
!
2
Note that number of ways in which 30 Jawans can be divided into 3 equal
groups = !
3
1
!
10
!
10
!
10
!
30
and the number of ways in which 30 Jawans can be deputed equally on 3 borders,
(10 on each border) is !
3
!
3
!
10
!
10
!
10
!
30
Remember that you have to divide by as many factorials as the number of equal groups.
e.g. number of ways in which 200 people can be divided into 100 couples is
(100 equal groups)
= !
100
.
)
!
2
(
!
)
200
(
100 .....(B)
Answer in (B) can also be written as
(i)
)!
100
.(
2
)
199
........
4
.
3
.
1
(
!
100
.
2
100
100
= 1.3.5 ....... 199.
(ii) !
100
.
2
)
200
).....(
102
)(
101
(
)!
100
(
100 = 

















2
200
.....
2
102
.
2
101
Consider the example : Number of ways in which 12 different books can be tied up
in 3 bundles one containing 2 and the other two bundles containing 5 different books.
Note that our final correct answer is !
2
!
5
!
5
!
2
!
)
12
(
...(1)
One way 
!
10
!
2
!
12
& 
!
2
!
5
!
5
!
10
 (1)
Second way =
!
7
!
5
!
12
& =
!
5
!
2
!
7

!
5
!
2
!
7
.
!
7
!
5
!
12
= !
2
!
5
!
5
!
12
seems to be a mistake of 2! in denominator. How do we explain.
Consider one group out of !
2
!
5
!
5
!
12
say ...(2)
Hence A B C D E | F G H I J | K L is one way.
However in (1) when we select F G H I J and reject the remaining i.e.
We have again the same group F G H I J | A B C D E | K L
Hence answer gives by (2) is double  correct answer is !
2
!
2
!
5
!
5
!
12

7. Number of games which can be played.
[ Method - 1 : !
3
!
2
!
2
!
2
!
6
·
3
= 45 ; Method - 2 : 6
C4
. 3 = 45 ]
(i)(b) 6 different books to be distributed between R | S | G if each child gets at least one book.
(Do it in two different ways)
Method 1: 36
– {3
C1
(26
– 2) + 3
C2
} ( note that : 6
C3
. 3! . 33
is wrong )
Method 2: 6 different things (123 , 114 and 222) – broad groups
= !
3
.
!
3
)
!
2
(
!
6
!
2
!
4
!
1
!
1
!
6
!
3
!
2
!
1
!
6
3









 ]
(ii) Number of ways in which 8 persons can be seated in 3 different taxies each having
3 seats for passangers and duly numbered if
(i) internal arrangement of persons inside the taxi is immaterial.
(ii) internal arrangement also matters.
[Sol. (i) The only way out is to accomodate the persons in groups of 3, 3 and 2.
No. of groups = !
2
!
2
!
3
!
3
!
8
& number of ways = 2
2
)
!
2
(
)
!
3
(
!
3
.
!
8
= 2
)
!
2
(
!
3
!
8
(ii) If internal arrangement matters then number of ways =9
C8
. 8! ]
(iii) 8 computers of different make to be distributed in 5 school if each school gets at least
one computer. (41111 ; 32111 , 22211)
(iv) During election 3 districts are to be convassed by 20, 15 and 10 people respectively. If
45 volunteers then the number of ways in which they can be sent. [ !
10
!
15
!
20
!
45
]
(v) In how many ways 13 cards to each of the four players be distributed from a pack of
52 cards so that each may have
(i) A | K | Q | J of the same suit ; (ii) A | K | Q | J of any suit
[Sol. (i) !
4
!
4
.
)
!
9
(
!
4
.
!
36
4
 ; (ii) In case of any suit
4
4
)
!
4
(
!
4
.
)
!
9
(
!
4
.
!
)
36
(
 ]
(vi) A f : A  B
Case-I: When both the sets A and B contain an equal number of elements
(i) Total number of functions = 44
= 256
(ii) Number of functions one-one = 4!
(iii) Number of functions many-one = 44
– 4!
(iv) Number of onto function 4!  number of bijective mapping = 4!
(v) Number of into functions = 44
– 4!
Alternate : 4
C1
[34
– {3
C2
+ 3
C1
(24
– 2)}] + 4
C2
(24
– 2) + 4
C3
· 1
Case-II: When number of elements in A(domain) is more than B
(i) Total number of functions = 45
= 1024
(ii) One-one injective = Nil
(iii) Many-one = 1024
(iv) Number of onto function = 







!
3
·
!
2
·
!
1
·
!
1
!
5
4! = 240
(v) Number of into function = 1024 – 240 = 784

8. (ii) Number of injective mapping = 5
C4
· 4! = 120
(iii) Number of many-one = 625 – 120 = 505
(iv) Number of onto function = 0
(v) Number of into function = 625
Home Work : Exercise-1 (Hall & Knight)
(7) Permutation of alike objects



time
a
at
some
taken
:
II
Case
time
a
at
all
taken
:
I
Case


(DADDY / MUMMY)
Number of permuation of n things
different
all
are
r
and
kind
another
of
q
kind
one
of
p





taken all at a time = !
r
!
q
!
p
!
n
For example number of words which can be formed using all letters of the word
“MAHABHARAT” = !
2
!
4
!
10
(A) Examples on All at a time
ASKING :
21 W and 19 B balls are arranged in a line (balls of the same colour alike). Find the
number of arrangement if Black balls are separated. (22
C19
= 22
C3
).
(a) ASSASSINATION | MISSISSIPPI | CONSTANTINOPLE | HORROR | HONOLULU |
Consider the word ASSASSINATION (Vowels - 6 ; Consonants- 7 ; | A | A | I | A | I | O | )
(i) no two vowels are consecutive
(ii) S’s are separated / no two ‘S’ occur together
(iii) without changing the order of vowels (AAIAIO only)
(iv) at least one ‘S’ is separated from the rest of the S’s ( Total – all four together)
(v) keeping the positions of each vowel fixed ( each V must remain at its own place)
(vi) without changing the relative order of vowels and consonants.
(b) Number of numbers greater than a million using the digits 2, 3, 0, 3, 4, 2, 5 of these
how many are div. by 5. [Ans. 1080]
(c) AA, BB, CC, D, E, G, F can be seated if persons of the same nationality are to be
separated. (47) 8!
(d) How many 8 digit numbers can be formed using two 1’s, two 2’s, two 3’s, one 4 and
one 5 so that no two consecutive digits are identical. [Ans : 2220]
(e) How manydifferent words can be formed using all the letters ofthe words HONOLULU
if (no two alike letters are together).
(Make Venn Diagram – 2220)
[Sol. Total numbers w/o any restriction = !
2
·
!
2
·
!
2
!
8
= 7 ! = 5040
n (E3
) = 45 = 5 !
——————
for n (E2
) : 33 45
n (E2
) = n (A  B) – n (A  B  C)
n (B  C) – n (A  B  C)
n (C  A) – n (A  B  C)

9. [Sol. Possible digits are 5, 6, 7, 8, 9
Category Number of 7 digits number Divisible by11
9 9 9 9 9 9 5 !
6
!
7
= 7 ×
9 9 9 9 9 7 7 !
6
·
!
5
!
7
= 21 ×
9 9 9 9 9 6 8 !
5
!
7
= 42 × × × × !
3
!
4
!
2
!
3
= 12
9 9 9 9 8 8 7 !
2
·
!
4
!
7
= 105 !
3
!
4
· 3 ! = 24
9 9 9 8 8 8 8 !
4
·
!
3
!
7
= 35 !
3
!
4
· 1 ! = 4
210 40 ]
(g) Four faces of a tetrahedral dice are marked with 2, 3, 4, 5. The lowest face being
considered as the outcome. In how many ways a total of 30 can occur in 7 throws.
[Sol. 5555532 ; 5555442 ; 5555433 ; 5554443 ; 5544444 ] [ Ans: 413 ]
(B) Examples on some at the time
(a) INDEPENDENCE – 5 letters word – Selections and arrangements,
[E’s = 4; N’s = 3 ; D’s = 2 ; I’s = 1; P’s = 1 ; C’s = 1 ]
(b) Number of different collections of 7 letters that can be formed from the letters of the
word ACCOMMODATION. (Ans: 315)
A’s =2; C’s =2 ; O’s = 3; M’s =2 ; D’s = 1 ; T’s = 1 ; N’s = 1 ; I’s = 1
(c) Number of numbers greater than 1000 from the digits 1,1,2,3,4,0 taken 4 at a time.
(Ans: 159)
(d) How many 6 lettered word can be formed using the letters from the word
INTEGRATION if each word has 3 vowels and 3 consonants.
Asking : MISSISSIPPI ( selection of 5 letters)–Six cases [ Ans: 25 ]
Concept: Coeffecient of xp
yq
zr
in the expansion of (x + y+ z)p + q + r
= !
r
!
q
!
p
)!
r
q
p
( 

]
e.g. Coeffecient of x5
y4
z3
in (x + y + z)12
is !
3
!
4
!
5
!
12
and coeff. of x6
y3
z3
is !
3
!
3
!
6
!
12
]
Explanation : [x + y + z]12
= [x + (y + z)]12
Tr + 1
= 12
Cr
· x12 – r
· (y + z)r
put r = 7
T8
= 12
C7
· x5
· (y + z)7
= 12
C7
· x5
· 7
Cp
· y7 – p
· zp
put p = 3
= 12
C7
· 7
C3
· x5
· y4
· z3
Hence coefficient x5
y4
z3
is 12
C7
· 7
C3
= !
4
·
!
3
!
7
!
5
·
!
7
)!
12
(
= !
5
·
!
4
·
!
3
)!
12
(
Hence arrange 12 object of which 5 alike, 4 OA and 3 OA taken all at a time in
!
5
·
!
4
·
!
3
)!
12
(
ways

10. If as many more words as possible be formed using the letters of the word
“CIVILISATION” without changing the relative order of vowels and consonants.
[ 1
!
6
!
4
!
6

 = 21599]
(8) Circular Permutation
alike
are
some
When
different
are
objects
When
(A) All five are the same arrangement in a circle. Hence number of circular permutations of
(i) (a) n different things taken all at a time = (n–1)! ;
(b) taken r at a time = n
Cr
.(r–1)!
This distinguishes/discriminates/considers to be differenttheclockwiseand anticlockwise
arrangement.
(ii) If clockwise and anticlockwise arrangements are same then
2
)!
1
n
( 





The number of circular permutations of n different things.
Example
(i) (a) Number of ways in which 7 Americans and 7 British people can be seated on a
round table so that no two Americans are consecutive. (Ans: 6! . 7!)
(b) Number of circular permutations of n persons if two specified people are never
together.
(ii) (a) Number of ways in which 8 persons can be seated on a round table so that
“all shall not have the same neighbours in any two arrangement”.
(note that clockwise & anticlockwise arrangement are not to be distinguished)
[Ans:
2
)!
1
8
( 
= 2520 ]
(b) Number of ways in which 10 children can sit in a mery go round relative to one
another. [Ans: 9! (Here clockwise & anticlockwise arrangement is different]
(iii) Out of 10 flowers of different colours, how many different garlands can be made if
each garland consists of 6 flower of different colour. [ Ans: 10
C6
.
2
!
5
]
(iv) Number of ways in which 5B and 5G can be seated on a circle alternately if a particular
B1
and G1
are never adjacent to each other in any arrangement.
[Hint : Total 5! 4! – 3! 4! = (4!)2
[5 – 2] = 3(4!)2
= 1728 ]
Alternatively : 4 ! · (G1
can not come in adjacent places of B1
 for G1
only 3 places
 4 ! · (3) · 4! for remaining 4 girls ]
(v) (a) n people A1
,A2
, ........,An
sitting on a circle. Number of ways in which 3 people
can be selected if no two of them are consecutive.
[Solution : n
C3
– [no. of ways when all 3 consecutive
+ when exactly two consecutive]
n
C3
– [n + n (n – 4)] = n
C3
– (n2
– 3n) ]

11. [Sol. Step (1) Select the initial vertex say ‘1’ (in 15
C1
ways)
(2) Now 2 and 15 cannot be selected. From the
remaining vertices 3 to 14 (twelve) we have
to select 5 more for our hexagon.
(3) Symbolise the vertices to be taken by and the vertices
not be taken ( 7 in this case) by X X X X X X X.
(4) Identify the gaps between these crosses (8 in this case) and select
any five out of these gaps in 8
C5
ways.
(5) Serial number are to be alloted either to or to a ‘X’ whichever
comes earlier. In the present problem the vertices corresponding to
one selection are 3, 5, 8, 11, 13 and the hexagon as shown
(6) For each selection we therefore have a hexagon with two non consecutive
vertices
(7) Number of hexagons = 15
C1
× 8
C5
. However this particular hexagon 1, 3,
5, 8, 11, 13 will occur 6 times when we select the initial vertex as 3 or 5 or
8 or 11 or 13. Hence our answer is 6 times more.
(8) Required number of hexagons = 140
6
56
15
6
C
15 5
8




Ans]
[Alternative Solution of V(b) : (Similar to that in linear) :
As in linear, let us open this chain to have 1, 2, 3, .....13, 14, 15
O O O O O O to be selected ; | X | X | X | X | X | X | X | X | X | not to be selected
Number of ways = 10
C6
Required number of ways = 10
C6
– number of ways when 1 and 15 are included , since in
circular these become consecutive
Now if 1 and 15 are already selected , 2 and 14 cannot be taken. Remaining vertices are
3, 4, 5, ..... , 11, 12, 13 (11) ; O O O O (4) ; | X | X | X | X | X | X | X | ( 7)
Cases to be rejected = 8
C4
Required number of ways = 10
C6
– 8
C4
 10
C4
– 8
C4
 210 – 70 = 140 ]
(vi) AA, BB, CC, D, E , G , F can be sealed on a circle if people of the same nationality are
to be separated.
(viii) 16 knights are sitting on a round table.
(B) Circular permutation when objects are alike.
TERRORISM (In how many ways the letters of this word can be arranged in a circle)
Let R’s are considered to be different for a moment say R1
, R2
, R3
& Let no. of
arrangements be x. Hence

total
!
3
x
= (9 –1)! = 8!
!
3
!
8
x 


12. when ‘0’ is ‘1’ is ‘2’ is (2n–1) is ‘2n’ is
always taken always taken always taken always taken always taken
0 0 6n 1 1 6n–2 2 2 6n–4 2n–1 2n–1 2n+2 2n 2n 2n
0 1 6n–1 1 2 6n–3 2 3 6n–5 2n–1 2n 2n+1
0 2 6n–2 1 3 6n–4 
  2 3n–2 3n
0 3n 3n 1 3n–1 3n 2 3n–1 3n+1
(3n+1) ways (3n–1) ways (3n–2) ways 2 way 1 way
Total = 1 +(2+4)+(5+7)+(8+10) ................. ( (3n–1) + (3n+1) )
= 1 + 6 [ 1+2+3+ ....+n) = 1 + 6 +
2
)
1
n
(
n 
= 3n2
+ 3n + 1
Note: ‘O’ is always taken means all 6n flowers
on the same side of the extreme G
|||ly 1 is always taken means at least 1 different
flower between two successive G’s.
(9) Total number of combinations :
Means selectingat least one out ofn things.
1
)......
1
r
)(
1
q
)(
1
p
(
C
.....
C
C
1
2 n
n
2
n
1
n
n











different
are
all
remaining
and
alike
other
some
alike
some
different
all
are
thing
when
(i) Number of different ways it is possible to draw a sum of many with a rupee, a 50 P ,
a 25 P,a 10 P, 5 P , 3 P , 2 P and 1 P coin is 28
– 1.
(ii) Out of 3 different maths (M1
| M2
| M3
) , 4 different physics (P1
| P2
| P3
| P4
) and 5 diff.
chemistry (C1
| C2
| C3
| C4
| C5
) books, how many different selections/collections can
be made so that each selection consists of
(a) exactly one book on each subject (3
C1
· 4
C1
· 5
C1
= 60)
(b) at least one book on each subject (23
–1)(24
–1)(25
–1) = 7×15×31 = 3255
(c) taking at least one book = 23
.24
.25
– 1
(iii) Out of 2 Cocas , 3 Mangoes and 4 apples how many different selections of fruits can
be made if each selection has
(i)at least one fruit Case I : fruits of the same species are alike
(ii)at least one fruit of every species. Case II : different
(iv) A shopkeeper places before you n different books each having p copies. Number of
different selections (p+1)n
– 1.
Note that every natural number except 1 has atleast 2 divisors. If it has
exactly two divisors then it is called a prime. System of prime numbers begins
with 2. All primes except 2 are odd. All primes are odd but all odds are not
prime. A number having more than 2 divisors is called a composite. 2 is the
only even number which is not composite. A pair of natural numbers are said
to be relatively prime or coprime if their HCF is one. For two natural numbers
to be relatively it is not necessay that one or both should be prime. It is
possible that they both are composite but still coprime. eg. 4 and 25. Note
that 1 is neither prime nor composite however it is coprime with every other
natural number.A pair of primes are said to be twin if their non-negative

13. (b) Number of proper divisors = 120 – 2 (note that 1 and N are not the proper divisors)
(c) Number of odd divisors = no 2 should be taken. = 4.3.2 = 24
(d) Number of even divisors = at least one 2 should be taken = 4.4.3.2 = 96
(e) Number of divisors divisible by 5 = atleast one 5 must be taken = 5.4.2.2 = 80
(f) Number of divisors divisible by 10 = atleast one 5 and atleast one 2 must be taken
= 4.4.2.2 = 64
(g) sum of all the divisors = (20
+21
+22
+23
+24
) (30
+31
+32
+33
)(50
+51
+52
)(70
+71
)
(h) Number of ways in which N can be resolved as a product of two divisors
(N = pa
. qb
.........; p ,q are prime) =










square
perfect
a
is
N
if
2
1
.......
)
1
b
(
)
1
a
(
square
perfect
a
not
is
N
if
.........
)
1
b
(
)
1
a
(
2
1
(i) Number of ways in which Ncan be resolved as a product of 2 divisors which are relatively
prime = 2n–1
where n is the number of primes involved in the prime factorisation of N.
(10) Summation of numbers – (3 different ways)
(a) Sum of all the numbers greater then 10000 formed by the digits 1,3,5,7,9 no digit being
repeated
Method - 1 All possible numbers = 5! = 120
If one occupies the units place then total numbers = 24.
Hence 1 enjoys units place 24 times
|||ly
1 enjoys each place 24 times
Sum due to 1 = 1 × 24 ( 1+ 10 + 102
+ 103
+ 104
)
|||ly
Sum due to the
digit 3 = 3 × 24 ( 1 + 10 + 102
+ 103
+ 104
)
      
Required total sum = 24 ( 1 + 10 + 102
+ 103
+ 104
) ( 1+ 3 + 5 + 7 + 9 )
Method –2 In 1st
column there are twenty four 1’s , Twenty four 3’s & so on and their sum is
= 24 × 25 = 600
Hence add. in vertical column normally we get = 6666600
5th
2nd
1st
X X X X X
X X X X X
    
    
X X X X X
____________________________
666 6 6 0 0 = 6666600 ]
________________________________________________
Method – 3 Applicable only if the digits used are such that they have the same
common difference. (valid even if the digits are repeating)
Writing all the numbers in ascending order of magnitude
S = ( 13579 + 13597 + ........... + 97513 + 97531)
S = ( 13579 + 99531) + (13597 + 97513) + ....
= (111110) 60 time = 6666600 Ans
n

14. repeated
(b) Sum of all the numbers greater than 10000 formed by the digits 0,1,2,4,5 no digit being
repeated
[Ans: 24(0+1+2+4+5) (1+10+102
+103
+104
) – 6 (1+2+4+5)(1+10+102
+103
)
(c) Sum of all distinct four digit numbers that contain only the digits 1,2,3,4,5 each at most
once is given by
[ Ans: 6(2+3+4+5) + 6(1+3+4+5) + 6(1+2+4+5) + 6(1+2+3+5) +6(1+2+3+4)
= 6  [ 4 ( 1+2+3+4+5)] = 24(1+2+3+4+5). (1+10+102
+103
) ]
(d) Sum of all the numbers that can be formed using all the digits 2, 3, 3, 4, 4, 4 is :
(A*) 22222200 (B) 11111100 (C) 55555500 (D)20333280
[Hint:4 x 30 [x] + 3 x 20 [x] + 2 x 10 [x] where [x] = 1 + 10 + 102 + 103 + 104 + 105 ]
(11) Distribution of alike objects
(i) Number of ways in which n identical coins can be distributed between p persons, if
each person receiving none , one or more coins = )!
1
p
(
!
n
)!
1
p
n
(



(ii) Ashelfcontains 6 separate compartments.Number ofways in which 12 indistinguishable
(like) marbles can be placed in the compartments so that no compartment is empty.
[ Ans : 462] [ 11
C5
]
(iii) Number of natural solution of the equation x+y+z = 102 (101
C2
= 5050)
(iv) Number of different dissimilar terms in (x1
+x2
+...........+xn
)m
= 





1
n
m
0
....
0
0
0
....
00




= m+n–1
Cm
alternatively stated as: Number of terms in a complete homogeneous expression of
degree m in n variables
Ask : (x + y + z)11
= ?
(v) (a) Number of non negative integral solution of the inequality x+y+z+t < 30
[Hint: add one more beggar say W and find the solution of the equality x+y+z+t+W=30 ]
(b) Number of points with integral co-ordinates that lie inside a triangle whose
co-ordinates are (0, 0), (0, 102) and (102,0) is [Ans : 5050]
[ Hint : x + y < 102 x  N, y  N
i.e. x + y  101
give 1 to each x and y
hence x + y  99
equivalent to x + y + z = 99
 101
C2
=
2
100
.
101

2
99
0
0
0
0
....
0
0
0
0 

 
 
 = 5050]
(vi) In how many ways 30 marks be allotted to 8 questions if atleast 2 marks are to be given
to each question, assuming the marks are to be alloted in non negative integral values.
[ Ans: 





7
14
0
....
0
0
0
....
00 


= 21
C7
]
(vii) There are unlimited number of R,W,B and G balls.
(balls are all alike except for the colour)
(i) Number of ways in which a selection of 10 balls can be made

15. (i) false
selected
be
to
10
= C3
= 286 (ii) false
6
= C3
= 84
(viii) Aman has to buy 25 mangoes in four different varities buying at least 4 of each variety.
In how many ways can he plan his purchases, if mangoes of each variety are identical
and available in abundance. [ Ans: 12
C3
]
(ix) Number of ways in which K identical balls can be distributed in p different boxes if no
box remains empty. [ Ans: )!
1
p
(
)!
p
K
(
)!
1
K
(



]
(x) Number of distinct throws in a throw of n indistinguishable ( alike ) ordinary dice.
treat these six faces as beggars.






5
throws
n
0
...
0
0
0
....
000 


= !
5
!
n
)!
5
n
( 
= n+5
C5
= n + 5
Cn
(xi) Number of ways in which 2 apples, 3 oranges and 4 mangoes can be distributed
between R|S|G if each child receives none, one or more number of fruits.
(Fruits of the same species are alike). [ Ans: 4
C2
. 5
C2
. 6
C2
]
(xii) In how many ways 4 alike apples and 4 alike oranges can be distributed in six children
if each child gets at least one fruit.
[Sol. 111113 ; 111122
Case 1 : 1 fruit to each of 5 children and 3 fruits to one child
Case 2 : 1 fruit to each of 4 children and 2 fruits to each of two children.
Now Case 1: Case 2 :
(xiii) A supermarket offers ice creams in 10 different varieties. Ice creams of each variety are
identical. In how many ways 4 ice creams can be selected
(i) without any restriction ; (ii) at least two ice creams of the same variety are taken.
V1
=
V2
=
   
V10
=
[Sol. (i) Treat V1
V2
.....V10
as beggar

selected
be
to
0000 


9
0
...
0
0 


= 13
C4
(ii) 13
C4
= All four different + 2 alike & 2 different + 2 alike & 2 OA + 3alike
& 1 different + All four alike.
Required no. of ways = 13
C4
– all four different
= 13
C4
– 10
C4
]
(xiv) In how many ways 30 can be partitioned into 3 unequal natural parts. Number of ways
when equal parts are also counted.

16. r r r+1 r r–1
(i) n
Cr
is maximum if









odd
is
n
if
2
1
n
or
2
1
n
r
even
is
n
if
2
n
r
eg. Find the difference between the greatest value of 15
Cr
and 12
Cr
(ii) A man wants to give as many number of parties as he can out of his 20 friends
by calling the same number of people but not the same group. How many should
he call so that the number of parties is maximum.
(iii) There are 2n letters which are either a’s or b’s. Find the maximum number of
arrangement of these letters. (Maximise 2n
Cr
. Hence r = n ]
(iv) Find the greatest number of different groups that can be formed from the letters
of the word “BAROUCHE”, each group consisting of the same number of
letters. In how many of them will the letter ‘C’ occur
[Maximise 8
Cr  8
C4
. If C is always taken then the number of ways = 7
C3
]
(b) Grid Problem
Complete cartesian plane is partitioned by drawing line | | to x and y axis equidistant
apart like the lines on a chess board. Then the
(i) Number of ways in which an ant are can reach from (1, 1) to (4, 5) via shortest path.
paths
vertical
1)
(m
and
horizontal
1)
(n
be
will
e
then ther
lines
horizontal
m
and
rational
n
are
there
If
:
Note
C
3!
4!
7!
ways
of
number
required
Hence
.
paths
V
4
and
H
3
traverse
to
has
it
;
ant
the
of
travel
of
mode
the
be
may
Whatever
.
[Sol
3
7




(c) Dearrangement :
(i) n letters are to be kept in n directed envelopes number of ways in which they can be
plated if none of the letter goes into its own envelope is = n! 









!
n
1
)
1
(
.......
!
4
1
!
3
1
!
2
1 n
[Proof: Let the required number of arrangements be Qn
and Let n! = Pn
The total number of arrangements Pn
includes Qn
in which none of the letters are put
into the right envelopes.
Let us find the number of permutations in which only one letter is put into the right
envelope.
This is n . Qn–1
Similarly, the number of permutations in which only two letters are put into their

17. Pn
= Qn
+ n
C1
. Qn–1
+ n
C2
Qn–2
+ n
C3
. Qn–3
+ ...... + nQ1
+ Q0
This can be written in symbols as
Pn
= (Q + 1)n
where QK
, PK
replace QK
, PK
etc.
We can write this symbolic identity valid  x as
(P + x)n
= (Q + 1 + x)n
where P = Q + 1
replacing x by –1, we get
Qn
= (P – 1)n
= Pn
– n
C1
. Pn–1
. 1 + n
C2
Pn–2
– n
C3
Pn–3
+ ...... + (–1)n–1
n P1
+ (–1)n
reverting to Qn
, Pn
we have
Qn
= 
 

 

zero
1
n
1
n
n P
C
P 

+ n
C2
Pn–2
– n
C3
Pn–3
+ .... + (–1)n–1
n P1
+ (–1)n
i.e. Qn
= 






 







!
n
)
1
(
)!
1
n
(
)
1
(
.........
!
4
1
!
3
1
!
2
1
!
n
n
1
n
]
(1) Problem on match the column.
(2) Number of ways in which 13 card combination can be distributed to each of the
4 players if each player gets Ace and king of different suit.
[Ans.  
!
4
·
9
·
!
4
)
!
11
(
!
4
·
!
44
4 ]
(d) Coefficient Method : Note that Coeff. of xr
in (1 – x)–n
= n + r – 1
Cr
( N
n )
It gives the number of solutions.
Consider the Examples :
(i) In an exam. maximum marks for each of the 3 papers are 50 each and for the 4th
paper
100. Number of ways in which a person can secure 60% marks.
(ii) 3 different dice are thrown. Number of ways in which a total of k (9  k  14) can be
had.
(iii) Show that the number of different selections of 5 letters which can be made from five
a’s four b’s; three 3 c’s; two d’s and one e is 71.
[Sol. Coeff. of x5
= (1 + x + x2
+ x3
+ x4
+ x5
) (1 + x + ... + x4
) (1 + x + .. x3
) (1 + x + x2
) (1 + x)
=
x
1
x
1
.
x
1
x
1
.
x
1
x
1
.
x
1
x
1
.
x
1
x
1 2
3
4
5
6










= 5
4
3
2
)
x
1
(
)
x
x
x
1
(




(neglecting higher degree)
= coeff. of x5
in (1 – x)–5
– coeff. of x3
(1 – x)–3
– coeff. of x2
in (1 – x)–5
– coeff. of x in (1 – x)–5
= 9
C4
– (7
C3
+ 6
C2
+ 5
C1
)
= 126 – (35 + 15 + 5) = 126 – 53 = 71]
(iv) Prove that the number of combinations of n letters out of 3n letters of which n are ‘a’
and n are ‘b’ and the rest unlike is (n + 2) 2n – 1
[T/S]
[Sol : Coeff of xn
in (1 + x + x2
+ .... + xn
) (1 + x + x2
+ .... + xn
) (1 + x)n
or coeff. of xn
in  n
2
1
n
x
1
.
x
1
x
1










 
or coeff of xn
in (1 – x)–2
. (1 + x)n
n n n n 2 n n –2

18. 2S = (n +2) (C0
+ C1
+ ..... + Cn
)  S = (n + 2)2n –1
(v) A bag contains 6n tickets numbered from 0, 1, 2, .... 6n–1. In how many ways 3 tickets
can be selected so that the sum of the numbers shown on them is equal to 6n.
[Sol. Number of way = coeff. of x6n
in (1 + x + x2
+ ..... + x6n–1
)3
– R
where R = number of cases to be rejected
or coeff. of x6n
in
3
n
6
x
1
x
1










coeff. of x6n
in (1 – x)–3
= 6n+2
C2
now R =






















0
n
3
n
3
n
2
n
2
n
2
4
n
6
2
2
2
n
6
1
1






= (3n –1) ways. (excluding 2n, 2n, 2n)
But each of such cases repeated 3 times + one case 2n, 2n, 2n
 R = 3 (3n – 1) + 1
 required number of ways
!
3
1
)
1
n
3
(
3
C2
2
n
6




]
(vi) In a shooting competition a man can score 0, 2, 3, 4 or 5 points for each shot number
of ways in which he can secure 30 in 7 shot.
[Sol : 5555550 ; 5555532 ; 5555442, 5555433; 5554443, 5544444
(Now compute arrangement in each case)
or Coeff. of x30
in (1 + x2
+ x3
+ x4
+ x5
)7
in two different ways.]
(vii) Find the number of integers between 1 to 100000 if the sum of their digits is 15.
[Sol. (X X X X X )
coefficient of x15
in ( 1 + x + x2
+ x3
+ ... + x9
)5
= (19
C4
– 5.9
C4
) = 735 Ans ]
Note : There are five papers each of maximum 9 marks. Number of ways in which the total of
15 can be had in these five papers.