Mathematics
Session
Permutation & Combination - 2
Session Objectives
Session Objective
1.
Combination
2.
Circular Permutation
Combination
Combination  Selection
Selection from a, b, c
Select one
Select two
Selection
a,
b,
c,
Rejection
b,c,
a,c,
a,b,
Selection
a , b,
b,c,
c,a,
Rejection
c,
a,
b,
No. of ways = 3
Combination
Selection
Select three a, b, c
Rejection

No. of ways = 1
Number of selection of some from a group.
= Number of rejection of remaining.
Combination
Number of ways of selecting a group of
two student out of four for a trip to Goa.
S 1,
S 2,
S 3,
S
4
Selection
Select two
Rejection
S1 S2
S3 S4
S1 S3
S2 S4
S1 S4
S2 S3
S2 S3
S1 S4
S2 S4
S1 S3
S3 S4
S1 S2
6 ways.
Combination
Number of ways of selecting one group
Of two for Goa other for Agra
Combination
Selection and Arrangement of 3
alphabets from A, B, C, D.
x.3!  4P3
4
P3
4!
x

 4 C3  4 C1  4 C(4 3)
3!
1!3!
Selection
Arrangement
Rejection
A, B, C,
ABC, ACB, BAC, BCA, CAB, CBA
D
A, B, D,
ABD, ADB, BAD, BDA, DAB, DBA
C
B, C, D,
BCD, BDC, CBD, CDB, DBC, DCB
A
A, C, D,
ACD, ADC, CAD, CDA, DAC, DCA
B
Combination
Number of distinct elements = n (1,2,3, .. n).
Ways of rejecting r = nCr
Ways of rejecting rest (n – r) elements
= nCn-r
nC
r=
nC
n-r
Particular selection 1,3, … r elements
Arrangement r!
Total no. of arrangement = nCr.r! = nPr
n
Cr 
n
Pr
n!

r!
(n  r)!r !
r nr
n
Cn  r 
 n Cr  n Cnr
n!
n!

n  (n  r) !(n  r)! (n  r)!r!
Questions
Illustrative Problem
There are 5 man and 6 woman. How
many way one can select
(a) A committee of 5 person.
(b) A committee of 5 which consist exactly 3 man.
(c) A committee of 5 persons which consist at least 3 man.
Solution :
Man – 5
(a)
11
Woman – 6 Total - 11
11!
C5 
6!5!
(b) Select 3 man
 5 C3
Select 2 women  6 C2
5
C3  6 C2
Solution Cont.
(c) At least – 3 man
Man – 5
Woman - 6
Composition of Committee
Case Man
Woman
3
2
5C
3
x 6C2 = 150
4
1
5C
4
x 6C1 = 30
5
0
5C
5
x 6 C0 = 1
No. of Ways = 150 + 30+ 1 = 181.
Illustrative Problem
In how many ways, a committee of 4
person Can be selected out of 6 person
such that
(a) Mr. C is always there
(b) If A is there B must be there.
(c) A and B never be together.
Solution :
No. of persons - 6
Committee - 4
(a) Available persons – 5
Ways = 5C3
Persons to select - 3
(b) Case – 1 : ‘ A is there’ – ‘AB’ in Committee.
Available persons – 4
4C
Ways
=
2
Persons to select - 2
Solution Cont.
Case – 2 – ‘A is not there’ – B may
/may not be there
Available persons – 5
5C
Ways
=
4
Persons to select - 4
No. of person - 6
Person to Select - 4
(c) ‘AB’ never together = total no. of committee
- ‘AB’ always together.
Total no. of committee = 6C4 .
‘AB together in committee’ = 4C2
 No. of Ways = 6C4 – 4C2 = 9
Illustrative Problem
How many straight lines can be drawn
through 15 given points. when
(a) No. three are collinear
(b) Only five Points are collinear
Solution :
Through two given point and unique straight line
(a)
15
C2
(b) 5 point s Collinear
 5 C2 distinct line  Considered as one
Number of Straight line 
15
C2  5 C2  1
Illustrative Problem
Find the number of 4 digit numbers
that can be formed by 3 distinct digits
among 1,2,3,4,5
Solution :- No. of digits = 5
5 digit
Select 3 distinct
5C
3 digit
3
Select one which
will repeat
3
C1
4!
 No. of ways  C3  C1 
2!
5
Form 4 digit nos.
using these three
3
Number of digits
formed.
4!
2!
Illustrative Problem
In how many ways 9 students can be
seated both sides of a table having 5
seat on each side (non-distinguishable)
Circular Permutation
A,B,C,D – to be seated in a circular table
Total line arrangement = 4!
abcd
dabc cdab
a
bcda
4 linear arrangement
d
b
1 circular arrangement
c
4 linear Arrangement  1 circular arrangement
4 ! linear Arrangement 
4!
or (4  1)! circular arrangement.
4
No. of circular arrangement of n object = (n-1) !
Question
Illustrative Problem
In how many way 4 girls and 5 boys can
be seated around a circular table such that
(i) No. two girls sit together
(ii) All girls sit together
(iii) Only two girls sit together
Solution
(i) Boys – 5
Girls – 4
B1
B2
B5
Boys  (5  1)!  4!
B4
B3
Girls  5  4!
Solution Cont.
(ii) Boys – 5
4!
G’s
Girls – 4
B1
B2
B5
B4
B3
Solution Cont.
(iii) Boys – 5
Girls – 4
2!
B1
G3 G4
B2
B3
B5
G2
B4
G1
Invertible Circular Arrangement
Ex :- Garland, Neck less.
Clockwise and anticlockwise arrangement
considered as same
B
B
A
C
D
For n objects.
C
D
1
No. of arrangement = (n  1)!
2
ABCD  ADCB
(4  1)!
2
Question
Illustrative Problem
I have ten different color stones. In
How many way I can make a ring of five
stones
Solution :
Stone - 10
Step 1 :- choose 5
10
C5
Step 2 :- arrange circularly (Invertible)
1
1
(5  1)!   4!
2
2
Ans : 
10
C5 
1
4!
2
Sum of Digits
Find the sum of all three digit numbers
formed by 1, 2 and 3
100th place 10th place Unit place
1
2
3
1
3
2
2
1
3
2
3
1
3
1
2
3
12
2
12
1
12
Sum of digits same (12)
Each digits is repeated
same no. of times
=2=(3-1) !
All digits comes equal
no. of times
Sum of digits in each column = (1+2+3) x 2! = 12
Sum of Digits
100th place 10th place Unit place
a
12
b
12
c
12
= 100a +10b +c
= 100x12 +10x12 + 12
= 12 (102 + 10 + 1)
= 12 x 111 = 1332
Sum of all numbers =
(sum of all digits) x (No. of repetition in a
particular column) x (No. of 1’s as number of
digits present in the number
Class Test
Class Exercise - 1
In how many ways can 5 boys and
5 girls be seated in a row so that no
2 girls are together and at least
2 boys are together?
Solution
First the boys can be seated in 5p5 = 5! ways.
Each arrangement will create six gaps:
__ B __ B __ B __ B __ B
If the girls are seated in the gaps, no 2 girls will be
together. Girls can be seated in the gaps in 6p5 = 6! ways.
But if the girls are in the first five or the last five gaps, no
2 boys will be together. So the girls can be seated in 6! –
(5! + 5!) = 6! – 2 × 5! = 4 × 5!
Solution contd..
Thus, total arrangements possible are 4 x (5!)2
= 4 × 120 × 120
= 57600
Note: Under the given condition, more than 2
boys cannot sit together.
Class Exercise - 2
Find the sum of all numbers formed
using the digits 0, 2, 4, 7.
Solution
Required sum is 43.13.1111  42.13.111



 

10n  1
10n  2  1
nn 1.S.
 nn  2 .S.

10  1
10  1

 42.13  4444  111
= 16 × 13 × 4333
= 208 × 4333
= 901264


Class Exercise - 3
If all the letters of the word ‘SAHARA’
are arranged as in the dictionary, what
is the 100th word?
Solution
Arranging the letters alphabetically, we
have A, A, A, H, R, S.
Number of words starting with A:
5!
 60
2!
5!
 20
Number of words starting with H:
3!
5!
 20
Number of words starting with R:
3!
Thus, the last word starting with R will be the
100th word. This is clearly RSHAAA.
Class Exercise - 4
How many numbers can be formed
using the digits 3, 4, 5, 6, 5, 4, 3
such that the even digits occupy
the even places?
Solution
Even digits are 4, 6, 4.
These can be arranged in the even places in
3! ways = 3 ways.
2!
The remaining digits: 3, 5, 5, 3 can be arranged
4!
 6 ways.
in the remaining places in
2! 2!
Thus, the total number of ways = 3 × 6 = 18 ways.
Class Exercise - 5
Ten couples are to be seated around
a table. In how many ways can they
be seated so that no two neighbours
are of the same gender?
Solution
Let all the members of one gender be
seated around the table. This can be done
in (10 – 1)! ways. Once one gender is
seated, arrangement of other gender is no
longer a problem of circular permutation
(since the seats can be identified). Thus,
the second gender can be seated in 10!
ways.
Thus, total ways = 9! × 10!
Class Exercise - 6
In how many ways can 15 delegates
be seated around a pentagonal table
having 3 chairs at each edge?
Solution
14
15 1
13
2
13
3
12
4
11
14 15
I
12
2
11
3
10
4
5
9
10
9
8
7
6
8
7
6
5
If we consider the problem as one of circular permutations, the
answer is (15 – 1)! = 14!
But we are considering the above two arrangements as same
while they are clearly different. All that has been done is that
all delegates have shifted one position. One move shift will also
give a new arrangement.
Solution contd..
But after three shifts, the arrangement will be
11
12 13
14
10
15
9
1
8
2
7
6
5
4
3
which is identical to the original arrangement.
Thus, we are counting three different arrangements as one.
Thus, number of actual arrangements possible = 3 × 14!
or 15! , where 5 is the number of sides of regular polygon.
5
Class Exercise - 7
Prove that the product of r consecutive
integers is divisible by r!
Solution
Let the r consecutive integers be
(n + 1), (n + 2), (n + 3), ..., (n + r)
Product = (n + 1)(n + 2)(n + 3) ... (n + r)

n!  n  1 n  2  n  3  ... n  r 
n!
n  r !


nr P
n!
But
n+rP
r
r
is an integer.
Thus, the product of r consecutive integers is divisible by r!
(Proved)
Class Exercise - 8
n
n
If n p  np
and
C

Cr 1,
r
r 1
r
find the values of n and r.
Solution
n
pr  npr 1

n!
n!

n  r  1
n

r
!
n

r

1
!
  

Also nCr  nCr 1
n!
n!


n  r ! r! n  r  1! r  1!

n!
n!

n  r ! r! n  r  1! r  1!
 r  2, n  3
Class Exercise - 9
A person wishes to make up as many
parties as he can out of his 18 friends
such that each party consists of the
same numbers of persons. How many
friends should he invite?
Solution
Let the person invite r friends. This can
be done in 18Cr ways. To maximise the
number of parties, we have to take the
largest value of 18Cr. When n is even,
nC will be maximum when r= n/2.
r
Thus, he should invite 18/2 = 9 friends.
Class Exercise - 10
In how many ways can a cricket team
of 5 batsmen, 3 all-rounders, 2 bowlers
and 1 wicket keeper be selected from
19 players including 7 batsmen,
6 all-rounders, 3 bowlers and
3 wicket keepers?
Solution
The batsmen can be selected in 7C5 = 21 ways.
The all-rounders can be selected in 6C3 = 20 ways.
The bowlers can be selected in 3C2 = 3 ways.
The wicketkeeper can be selected in 3C1 = 3 ways.
Thus, the total ways of selecting the team
= 21 × 20 × 3 × 3 = 3780 ways.
7.6
Note: 7 C5  7C2 
 21
2.1
Class Exercise - 11
In how many ways can we select one
or more items out of a, a, a, b, c, d, e?
Solution
We can select ‘a’s in 0 or 1 or 2 or 3, i.e. in 4
ways.
We can select ‘b’ in 2 ways, i.e. either we
select it or we do not select it and so on.
 The required number of ways in which
we can select one or more items is
(3 + 1)(1 + 1)(1 + 1)(1 + 1)(1 + 1) – 1
 4.24  1  63 ways.
Class Exercise - 12
In how many ways can we divide
10 persons
(i) into groups of 5 each,
(ii) (ii) into groups of 4, 4 and 2?
Solution
10
(i)
C5
2!
We have divided by 2!, because if we interchange persons
in group one, with persons in group two, the division is
not different, i.e.
group 1
group 2
group 2
group 1
abdfj
ceghi
ceghi
abdfj
10
(ii)
C4 .6 C4 .2 C2
10!
1


2!
4! 4! 2! 2!
Thank you