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Permutations where some objects are repeated
Consider the number of permutations of the letter
DEFEATED
There are three E’s and two D’s.
The number of permutations of D1 E1 F E2 A T E3 D2
But E1, E2, E3 can be arranged in
D1 D2
can be arranged in
is 8!
3! ways, and
2! ways, so
The number of arrangements of D1 E1 F E2 A T E3 D2 is
3! x 2! times the number of arrangements of D E F E A T E D.
Therefore the number of permutations of D E F E A T E D is
8!
 3360
3! 2 !
Generalising this argument we see that,
The number of permutations of n objects comprising of
r1 identical objects, r2 identical objects, ..……….,
rk identical objects is
n!
r1 ! r2 !..........rk !
Example
How many different permutations can be made using the
letters of the words
(i) BOOKS
(ii) LOTTO
(iii) MATHEMATICS
Solution
(i) BOOKS
n=5
n!

r!
r1 = 2
5!
 (5)(4)(3)  60
2!
The number of different permutations is 60.
(ii) LOTTO
n = n5!
r1 = 25!
r2 = 25  4  3 


r1 ! r2 !
2! 2!
2
There are 30 different arrangement.
 30
Example
There are 2 copies of each of 3 different books to be
arranged on a shelf. In how many distinguishable ways can
this be done?
Solution
n = 2 x 3 = 6 ( there are six books )
r1 = 2
r2 = 2
n!

r1 ! r2 ! r3 !
r3 = 2
6!

2! 2! 2!
 6  5  4  3 
 2  2 
 90
There are 90 ways to arrange 2 copies of each of 3 different
books on a shelf.
Example
How many different 10-letter codes can be made using three
a’s, four b’s, and three c’s?
Solution
n = 10
r1 = 3
r2 = 4
r3 = 3
10!

The number of such codes is
3!4!3!
 10  9  8  7  6  5 
 3  2  3  2 
= 4200.
Example:
In how many ways can 3 red, 4 blue and 2 green pens be
distributed among nine students seated in a row if each
student receives one pen?
Solution
The NOP =
9!
3! 4! 2!
= 1260 ways
Example
In how many of the possible permutations of the letters of the
word ADDING are the two D’s:
(i) together,
(ii) Separated
Solution
(i) There are five different items ( A, (DD), I, N, G ) which can be
arranged in 5! ways.
The number of possible permutations is 5! = 120.
(ii) D’s separated = without restriction - D’s together.
6!
2!
=
=
240.
-
5!
Example
How many different arrangements are there for the letters of
the word ARRANGEMENTS if
it begins with “R” and ends with “E”
the two letters “E” are separated
the two letters “E” and the two letters “A” are together
the consonant letters G, M, T, S are together
the two letters “N” occupied both ends
n (letters) = 12
n(A) = 2, n(R) = 2
Solution
n(N) = 2, n(E) = 2
R
E
a)
a)
b)
c)
d)
e)
1 way
10 letters
The NOP = 1 
1 way
10!
1
2! 2!
(2! for A, 2! for N)
b) the two letters “E” are separated
The NOP separated = without restriction – E’s together
Without restriction
=
12!
2! 2! 2! 2!
E’s together
=
11!
2! 2! 2!
EE
10 letters
Thus, the NOP in which the two E’s separated
=
12!
2! 2! 2! 2!
-
11!
2! 2! 2!
(2! for A, 2! for R,
2! for N)
c) the two letters “E” and the two letters “A” are together
EE
AA
The NOP =
8 letters
10!
11
2! 2!
d) the consonant letters G, M, T, S are together
GMTS
The NOP =
8 letters
9!
2! 2! 2! 2!
 4!
e) the two letters “N” occupied both ends
N
N
10 letters
The NOP =
1 
10!
2! 2! 2!
1
Exercise
A dancing contest has 11 competitors, of whom three are
Americans, two are Malaysians, three are Indonesians, and
three are Italians. If the contest result lists only the
nationality of the dancers, how many outcomes are possible?
Combinations of a set of objects
• A combination is a selection of objects with no consideration
given to the order (arrangement) of the object.
• So while ABC and BCA are different permutations they are
the same combination of letters.
• PQR, PRQ, QPR, QRP, RPQ, RQP are considered as
1 combination (because the order is not considered) and
6 permutation (because the order is considered).
Example 1:
Determine whether each of the following is a permutation or
combination:
a) 5 pictures placed in a row.
(Permutation)
b) 3 story books picked from a rack.
(Combination)
c) A team of 9 players chosen from a group of 20.
(Combination)
d) The arrangements of the letters in the word OCTOBER.
(Permutation)
e) Types of food in a plate taken for lunch consist of rice,
vegetables, chicken curry and prawn paste sambal.
(Combination)
f) Answering questions for Mathematics paper I.
(Permutation)
Therefore, we can conclude that permutations are used when
order is important and combinations are used when order is
not important.
Combinations of r object from n objects
3 students could be chosen, in order, from a total of 18 in
18 !
ways.
15 !
However, each of these choices can be arranged in 3! ways (ABC,
ACB, BAC, BCA, CAB, CBA),
so the number of combinations of three items chosen from 18 is
18!
= 816.
15! 3!
The general result is:
The number of combinations (or selections) of r objects
chosen from n unlike objects is
n
n!
Cr 
n  r  ! r !
Consider the following table;
n
taken r Combination
different object n C  n !
r
n  r  ! r !
object
The NOC of 3 objects A, B, C
taken 2 at a time = 3C2 = 3
NOC
Permutation
n!
n
P r
n  r  !
NOP
A, B
2
AB
1
AB, BA
2
A, B, C
2
AB, AC, BC
3
AB, BA, AC,
CA, BC, CB
6
1
ABC, ACB,
BCA, BAC,
CAB,CBA
6
A, B, C
3
ABC
Thus, we can see that the NOC is always less than the NOP.
Example 2
A quiz team of four is chosen from a group of 15 students. In how
many ways could the team be chosen?
Solution
15
C
4
15!

 1365
11! 4 !
Therefore the team can be chosen in 1365 ways.
Example 4
A school committee consists of six girls and four boys. A social
sub-committee consisting of four students is to be formed. In how
many ways could the group be chosen if there are to be more girls
than boys in the group?
Solution
If there are to be more girls than boys in the group then the group
will either have
four girls and no boys
Four girls can be chosen in
or
6
C4
three girls and one boy
= 15 ways
Three girls and one boy can be chosen in
6
C3
x
4
C1 = 80 ways
Therefore the number of ways of choosing the group if there are
more girls than boys is
15 + 80 = 95 ways
Example 3
If there are eight girls and seven boys in a class, in how many
ways could a group be chosen so that there are two boys and two
girls in the group?
Solution
8
Two girls can be chosen in
C
2
7
Two boys can be chosen in
C
2
ways
ways
Using the multiplication principle , number of ways of selecting
the group
8
7
 C2  C2

8!
7!

6 ! 2 ! 5! 2 !
=
28 x 21
= 588
Example 5
Given the set S = {a, b, c, d, e} consists of 5 elements. List all
the subsets of S with
(a) two elements
(b) four elements
a) {ab}, {ac}, {ad}, {ae}, {bc}, {bd}, {be}, {cd}, {ce}, {de}
b) {abcd}, {abce}, {abde}, {acde}, {bcde}
Example 6
In a football training squad of 24 people, 3 are goalkeepers, 7 are
defenders, 6 are midfielders and 8 are forwards. A final squad of 16
selected for a match must consist of 2 goalkeepers, 4 defenders, 5
midfielders and 5 forwards. Find the number of possible selections
if one particular goalkeeper, 2 particular defenders, 3 particular
midfielders and 3 particular forwards are automatically selected.
Solution
Number of ways of selecting the goalkeepers = 1C  2C  2
1
1
Number of ways of selecting the defenders = 2C  5C  10
2
2
3
C 3  3C 2  3
Number of ways of selecting the midfielders =
Number of ways of selecting the forwards =
3
C 3  5C 2  10
Number of ways of selecting the squad
= 2 x 10 x 3 x 10
(by using the principle of multiplication)
= 600
Example 7
ABCDEFGH is a regular octagon.
(a) How many triangles can be formed with
the vertices of the octagon as vertices ?
(b) How many diagonals can be drawn by
joining the vertices?
A
B
H
C
G
Solution
F
(a) A triangle is formed by taking 3 of the vertices.
Number of triangles = 8C3  56
(b) A line can be formed by taking any 2 points from the 8
vertices of the octagon.
8
Number of lines formed = C2  28
These 28 lines include the 8 sides of the octagon
Thus the number of diagonals = 28 – 8
= 20
D
E
Example 8
15 students are divided into 3 groups, with A having 7
students, group B having 5 students and group C having 3
students. Find the number of ways to form
a) the 3 groups
b) the 3 groups with 2 given students must be in group A.
Solution
a) The NOC =
15C
7
b) The NOC = (2C2
x 8C5 x 3C3 = 360360
x 13C5 ) x 8C5 x 3C3 = 72072
Example 9
A 3 member committee is to be formed from 4 couples of
husband and wife. Find the possible number of committees
that can be formed if
a) all the members are men
b) the husband and the wife cannot be in the committee at
the same time.
Solution
a) The NOC =
4C
3
= 4 ways.
b) The number of selecting 3 groups from 4 groups =
4C
3
The number of choosing a person from each of the 3
groups selected
= 2C1 x 2C1 x 2C1
=8
Thus, the NOC
=
4C
3
x 2C1 x 2C1 x 2C1 = 32 ways
Example 10
In a test, a candidate is required to answer 8 out of 10
questions. Find the number of ways a candidates
a) can answer the questions
b) can answer the question if the first 3 questions must be
answered.
Solution
a) The NOC =
b) The NOC =
10
3
C8
= 45
7
C 3x C 5
= 21
Example 11
In how many ways can a teacher choose one or more
students as a prefects from 5 eligible students?
Solution
The NOC is may be one or two or three or four or five person
chosen
=
5
C1
= 66
+
5
C2
+
5
C3
+
5
C4
+
5
C5