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1. Permutation & Combination
Permutation & Combination
Fundamental principle of counting :
If an event can occur in m different
ways following which another event can
occur in n different ways, then total
number of simultaneous occurrences
of both the events in a definite order
is (m × n). This can be extended to any
number of events.
Eg: For an event to be occur in m
1
, m
2
…, m
n
ways then number of ways is
m
1
× m
2
…m
n
Model-I :
Find number of ways of in which one
can travel from T
1
(town 1) to T
3
(town 3)
via T
2
(town 2).
Total ways :
It is easy to proceed by FPC
T
1
to T
2
T
2
to T
3
Total ways = 3 × 2 = 6
Model-II :
(i) To find the number of ways by which a
person can enter and leave cinema hall
by a different door.
Point to Remember!!!
FPC (Fundamental Principle of
Counting) is used to count some
event without actually counting
them.
Let us take help of some model.
Permutation & Combination 2.
Q.
How many 3 digit numbers can be formed by the digit 1, 2, 3, 4, 5 without
repetition.
Sol.
Hundred’s place digit can be selected in 5 ways.
Ten’s place digit can be selected in 4 ways.
Unit’s place digit can be selected in 3 ways.
So, 5 × 4 × 3 = 60
By Fundamental Principle of Counting,
A person can enter in cinema hall by 5 ways
and leave by 4 ways = 5 × 4 = 20.
(ii) If he can enter and leave by any door then
number of ways = 5 × 5 = 25
Basic Steps to Remember :
Step-I :
Identify the independent events involved in
a given problem.
Step-II :
Find the number of ways performing/
occurring each event
Step-III :
Multiply these numbers to get the total
number of ways of performing/occurring all
the events.
Permutation & Combination 4.
Q.
How many 6 digits odd number greater than 6,00,000 can be formed from the
digits 5, 6, 7, 8, 9, 0 if repetition of digit is allowed?
Sol.
Numbers greater than 6,00,000 and formed with the digit 5, 6, 7, 8, 9, 0 are of 6
digit but begin with 6, 7, 8 or 9.
Also, the numbers which end with 5, 7, 9 are odd.
Hence, first place can be filled by 4 ways (out of 6, 7, 8 or 9). Last place can
be filled by 3 ways. Hence, first and last place can be filled by 4 × 3 ways.
Also 2
nd
place can be filled by 6 ways.
rd
place can be filled by 6 ways.
th
place can be filled by 6 ways.
th
place can be filled by 6 ways.
Hence, all the 6 places can be filled by
4 × 3 × 6 × 6 × 6 × 6 = 15552 ways.
Q.
How many integers greater than 5000 can be formed with the digit 7, 6, 5, 4,
and 3 using each digit at most once.
Sol.
For 4 digits number ⇒ First position can be filled by 7, 6 or 5 (that is in three
ways). Hence
4 digit number = 3 × 4 × 3 × 2 = 72
5 digit number = 5 × 4 × 3 × 2 × 1 = 120
Total numbers = 192
Q.
How many natural numbers less than 30000 can be made from the digits 0, 1,
Sol.
Let a five digit number be denoted by
a b c d e
Each of the places can be filled by either of 0, 1, 2, 3, 4, 5, 6 in 7 ways.
The place marked by “a” can be filled by the digits 0, 1 or 2 (since number is
to be less than 30000)
Hence number of integers
4
= 3 × 7 × 7 × 7 × 7 = 3 × 7 In these numbers one
case includes “00000” which is not a natural number.
Hence number of natural numbers =3 × 7
4
5. Permutation & Combination
Q.
Consider the word DAUGHTER. How many 4 letter word can be formed from
the letters of above word so that each word contain letter G.
Sol.
4 possible positions for G.
Remaining three by ⇒ 4 × 7 × 6 × 5 = 28 × 30 = 840
Alternative method :
Total number of ways by which 4 letter word can be formed = 8 × 7 × 6 × 5
Number of four letter word without G = 7 × 6 × 5 × 4
Required ways = 8 × 7 × 6 × 5 – 7 × 6 × 5 × 4
Q.
How many different words can be formed using all the letters in the word
“MIRACLE”.
(a) If vowels may occupy the even position.
(b) If vowels may occupy odd position.
Sol.
(a)
vowels → I, E, A
consonants → M, R, C, L
Three vowels at three position ⇒ 3 × 2 × 1 = 6
Four consonants at four position ⇒ 4 × 3 × 2 × 1 = 24
Total number of ways = 6 × 24 = 144.
(b)
st
vowel have 4 choice
nd
vowel have 3 choice
rd
vowel have 2 choice
Thus total ways = (4 × 3 × 2) × (4 × 3 × 2 × 1) = 576
Q.
There are m men and n monkey. Number of ways in which every monkey has a
master, if a man can have any number of monkey.
Sol.
Monkey is distributed among masters, like 1 monkey can go to ‘m’ masters
Total number of ways = m × m × … × m = m
n
7. Permutation & Combination
Lexicography illustrations
Q.
Find total number of 5 letter word that can be formed from letters of word
“TOUGH”
Sol.
Q.
Find the rank of “TOUGH” if all the letters of the word are arranged in all pos-
sible orders and written out as in a dictionary.
Sol.
The number of letters in the word “TOUGH” is 5 and all the five letters are
different.
Alphabetical order of all the letters is G, H, O, T, U
Number of words beginning with G = 4 × 3 × 2 × 1
Number of words beginning with H = 4 × 3 × 2 × 1
Number of words beginning with O = 4 × 3 × 2 × 1
Number of words beginning with TG = 3 × 2 × 1
Number of words beginning with TH = 3 × 2 × 1
Number of words beginning with TOG = 2 × 1
Number of words beginning with TOH = 2 × 1
Next words beginning with “TOU” and it is “TOUGH” = 1.
Rank = 24 + 24 + 24 + 6 + 6 + 2 + 2 + 1 = 89
Q.
If the letters of the word “VARUN” are written in all possible ways and then
are arranged as in a dictionary, then the rank of the word VARUN is:
(A) 98 (B) 99 (C) 100 (D) 101
Sol.
(C)
VARUN
Total number of 5 letter words = 5! = 120
Number of words starting with A = 4 × 3 × 2 × 1 = 24
Number of words starting with N = 4 × 3 × 2 × 1 = 24
Number of words starting with R = 4 × 3 × 2 × 1 = 24
Know the facts
Lexicography is called science of
making words.
Permutation & Combination 8.
Number of words starting with U = 4 × 3 × 2 × 1 = 24
When V is fixed at first place, A on second place
and N on third place = 2 × 1 = 2
R on third place, N on fourth place = 1
R on third place, U on fourth placed and N on 5
th
place. This is our word.
Rank = 24 + 24 + 24 + 24 + 2 + 1 + 1
Q.
How many natural numbers are their from 1 to 1000 which have none of their
digits repeated.
Sol.
One digit natural numbers = 9
Two digit natural numbers = 9 × 9 = 81
Three digit natural number = 9 × 9 × 8 = 648
Total = 738
Q.
A man has 3 jackets, 10 shirts, and 5 pairs of slacks. If an outfit consists of a
jacket, a shirt, and a pair of slacks, how many different outfits can the man
make?
Sol.
3 Jackets, 10 shirts, 5 pair of slacks
Total ways = 3 × 10 × 5
Q.
There are 6 roads between A and B and 4 roads between B and C.
(i) In how many ways can one drive from A to C by way of B?
(ii) In how many ways can one drive from A to C and back to A, passing
through B on both trips?
(iii) In how many ways can one drive the circular trip described in (ii) without
using the same road more than once.
Sol.
6 4
A → B →C
(i) 6 × 4 = 24
(ii) 6 × 4 × 4 × 6 = 576
(iii) 6 × 4 × 3 × 5 = 360
Permutation & Combination 10.
Q.
If repetitions are not permitted
(i) How many 3-digit numbers can be formed from the six digits
2, 3, 5, 6, 7 and 9?
(ii) How many of these are less than 400?
(iii) How many are even?
(iv) How many are odd?
(v) How many are multiples of 5?
Sol.
(i) 6 5 4 = 120
(ii) 5 4 or 5 4
2 3
↑ ↑
= 20 + 20 = 40 ways
(iii) 4 5 2
= 40 ways
(iv) 5 4 4
= 80 ways
(v) 4 5 1
5
↑
= 20 ways
Q.
It is required to seat 5 men and 4 women in a row so that the women occupy
the even places. How many such arrangements are possible?
Sol.
M W M W M W M W M
= 5 × 4 × 3 × 2 × 1 × 4 × 3 × 2 × 1 = 2880
11. Permutation & Combination
NOTATION OF FACTORIAL
Notation of factorial and its Algebra :
The continued product of first n, natural
number is called as “n factorial” and denoted
by n! or n.
n! = 1 ∙ 2 ∙ 3 ∙ 4 ∙ … (n–1) ∙ n
Special Results
- 0! = 1 i.e. factorial of zero is 1.
Proof :
n! = (n–1)! ∙ n
Putting n = 1
- Factorial of negative number is undefined
(n–1)! =
n!
n
if n = 0 then (–1)!
= Not defined.
Q.
Find n if
(i) (n+ 1)! = 12 × (n – 1)! (ii) (n+ 2)! = 2550 n!
Sol.
(i) (n + 1)n (n–1)! = 12 × (n–1)!
n
2
(n + 4) (n – 3) = 0 ∴ n = 3
(ii) (n + 2) (n + 1) = 2550;
(n + 52) (n – 49) = 0 ∴ n = 49
n
n! [1 ∙ 3 ∙ 5 … ∙ (2n – 1)]
Proof :
(2n)! = 1 ∙ 2 ∙ 3 ∙ 4 ∙ 5 ∙ 6 ∙ … (2n – 1) (2n)
Take 2 common each n even terms
n
(1 ∙ 3 ∙ 5 … (2n - 1)) (1 ∙ 2 ∙ 3 … n)
n
(n!) (1 ∙ 3 ∙ 5 … ∙ (2n - 1))
13. Permutation & Combination
Q.
Find exponent of prime 2 in (100!)
Sol.
(50)
∙ 50! (1 ∙ 3 ∙ 5 … 99) → 50 2’s
(25)
∙ 25! (1 ∙ 3 ∙ 5 … 49) → 25 2’s
(12)
∙ 12! (1 ∙ 3 ∙ 5 … 23) → 12 2’s
(6)
∙ 6! (1 ∙ 3 ∙ 5 …11) → 6 2’s
(3)
3! (1 ∙ 3 ∙ 5) → 3 2’s
(1)
∙ (1 ∙ 3) → 1 2’s
Alternative Method :
2
2 3 4 5 6
E 100!
Q.
Find number of zeros at the end of (1000)!
Sol.
In any usual factorial of a natural number of 2s are more than number of 5s.
Hence number of 10s are same as number of 5s.
Basic Method :
Now we decide number of 5s in (1000)!
⇒ 200 number containing at least one 5
⇒ 40 number containing at least two 5
⇒ 8 number containing at least three 5
⇒ 1 number containing at least four 5
Total = 200 + 40 + 8 + 1 = 249
Objective approach :
5
2 3 4
E 1000!
Point to Remember!!!
Such kind of counting is possible
only on the basis of primes.
Permutation & Combination 14.
Q.
Find number of zeros at the end of 2007!
Sol.
5 2 3 4
E 2007!
Q.
Find exponent of 3 in 100!
Sol.
Basic method :
One 3 3, 6, 9, 12 … 99 = 33 ⇒ 33 number containing at least one 3
Two 3 9, 18, 27 … 99 = 11 ⇒ 11 number containing at least two 3
Third 3s 27, 54, 81 = 3 ⇒ 3 number containing at least three 3
4s 81 = 1 ⇒ 1 number containing at least four 3
Sum = 33 + 11 + 3 + 1 = 48
Objective approach :
5
2 3 4
E 100!
Q.
Find the exponent of 80 in 200!
Sol.
4
× 5
To find the exponent of 80 in 200!, we find the exponent of 2 and 5.
Exponent of 2 is
2 3 4 5 6 7
Exponent of 5 is
2 3
Now, exponent of 16 in 200! Is [197/4] = 49.
Hence, exponent of 80 is 49.
Q.
Find the number of zeroes at the end in product 5
6
7
8
9
10
31
Sol.
Given product P = 5
6
7
8
9
31
= [(5)
6
11
16
21
26
31
] λ (where λ is an integer)
∴ Exponent of 5 in P = 6 + 11 + 16 + 21 + 2 (26) + 31 = 137
∴ Number of zeros = 137
Permutation & Combination 16.
Combination :
Combination/selection/collection/committee
refers to the situation where order of
occurrence of the event is not important.
Combination is selection of one or more
things out of n things which may be alike or
different taken some or all at a time.
Example :
(i) Out of A, B, C, D take 3 letters and form
number plate of car. [Permutation]
(ii) Out of four letters A, B, C, D take any 3 letters
and form triangle (possible). [Combination]
In 1
st
case arrangement of letters are there, in
nd
case only selection will form the triangle,
arrangement is not required.
Theorem related to application of Permutation
and combination :
Theorem-
Number of permutations of n distinct things
taken all at a time symbolised as :
( )
n n
n n
P = P n, n = A =n!
Proof :
Let these are n things arranged at n places
n ∙ (n–1) ∙ (n –2) …3 ∙ 2 ∙ 1 = n!
We also say that number of ways in which
n distinct object can be arranged amongst
themselves in
n
P
n
= n! i.e. Find total number
of words that of 10 letters that can be formed
from all the letters of word GANESHPURI.
A =
n
P
n
= 10! = n!
Theorem-
Number of permutations of n distinct things
taken r at a time
0 ≤ r < n
n
P
r
= P(n, r) = A
r
n
n!
n −r!
Point to Remember!!!
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.
Hence we say that permutation is
arrangement of things in definite
order.
17. Permutation & Combination
Theorem-
Number of combination/selection of n distinct
things taken r at a time
n
r
n n!
C C n,r
r r! n r!
Proof :
Let 10 different objects are given as A, B, C,
D, E, F, G, H, I, J
Let combinations taking 3 at a time = x
Arrangement = (x)(3!)
x ∙ 3! =
10
P
3
10
3
P
x
Theorem-
Number of combination of n different things
taken r at a time when p particular things are
always included. =
n–p
C
r–p
E.g.: Total number of ways of selecting 11
player out of 15 player when Mahendra Singh
Dhoni and Yuvraj Singh are always included
15-
C
11–
l
C
9
Things T
1
, T
2
, … T
n
Places 1, 2, 3 … r
Choice n ∙ (n – 1) ∙ (n - 2) ∙ … ∙ [n - (r - 1)]
Total ways =
n n 1 n 2 ... n r 1 n r!
n!
n r! n r!
Hence, we can say that
n
r
n!
P
n r!
=
−
Q.
In how many ways can 5 person be made to occupy
(a) Five different chairs
(b) Three different chairs
Sol.
(a)
5
P
5
(b)
5
P
3
= 5 x 4 x 3 = 60
19. Permutation & Combination
Sol.
(i) Number of lines
Method-I:
Given 4 points are collinear so total number
of ways of selecting any two points =
10
C
2
4 collinear points give only one line (AB). So
over counted number of lines formed by col-
linear points =
4
C
2
Thus, total lines =
10
C
2
4
C
2
Method-II:
Take any two points from upper arc =
6
C
2
ways.
Take one point from upper arc and one point from line AB =
6
C
1
×
4
C
1
ways.
Take both the points from line AB then number of lines = 1 (Line AB).
Total number of lines =
6
C
1
×
4
C
1
6
C
2
(ii) Method-I :
Number of ways of taking any three points =
10
C
3
Number of ways of taking any three points from four collinear points =
4
C
3
Number of triangles formed =
10
C
3
4
C
3
Method-II :
Two points from upper arc and one point from line AB =
6
C
2
x
4
C
1
Two points from line AB and one point from upper arc =
6
C
1
x
4
C
2
All the three points from upper arc =
6
C
3
Total number of triangle =
6
C
2
×
4
C
1
6
C
1
×
4
C
2
6
C
3
Q.
A regular polygon of 10 sides is constructed. Triangles are formed joining
vertices of the polygon. Find the number of triangles
(i) if two sides of triangle coincide with the sides of polygon.
(ii) if only one side of triangle coincide with the side of polygon.
Sol.
We have regular polygon of 10 sides.
Triangles are formed joining vertices of
this polygon.
(i) Two sides of triangle coincide with
the sides of polygon. This is possible
only if three consecutive vertices of
polygon are selected as shown in the
following figure.
Permutation & Combination 20.
So, we have triangle A
1
A
2
A
3
, A
2
A
3
A
4
A
8
A
9
A
10
Thus, 8 such triangles are possible.
(ii) Only one side of triangle coincide with
sides of polygon.
Consider triangles in which one side
is A
1
A
2
Clearly third vertex cannot be A
3
or A
10
(otherwise two sides of triangle coin-
cide with the side of polygon)
So, for third vertex we have only six
choices (A
4
, A
5
, …, A
9
Thus, number of triangles with one
side A
1
A
2
is six.
Similarly, for each side of polygon
there will be six triangles.
So, number of triangles is 10 × 6 = 60.
Q.
Find the number of diagonals in the convex polygon of n sides.
Sol.
Diagonal of the polygon is formed, if two non-consecutive vertices of polygon
are joined.
So, number of diagonals = numbers of ways we can select two non-
consecutive vertices of polygon.
The first vertex can be selected in n ways.
Let A
1
is chosen first.
Now diagonal cannot be formed if any of A
2
and A
n
is chosen.
Hence for A
1
another vertex can be selected in (n – 3) ways from remaining
(n – 3) vertices.
Similarly, there are (n – 3) diagonals passing through each vertex.
So, by multiplication principle of counting,
Number of diagonals = n × (n – 3)
But, in above answer there is double counting.
Let A
1
is chosen as first vertex, then against it sometimes A
4
is chosen as
second vertex.
Similarly, when A
4
is chosen as first vertex, then against it sometimes A
1
is
chosen as second vertex.
Thus, each pair of vertices is selected twice.
Hence total number of diagonals =
n(n 3)
