1. 3! ways, and 2! ways, so Permutations where some objects are repeated
Consider the number of permutations of the letter
D E F E A T E D
There are E’s and D’s.
The number of permutations of D1 E1 F E2 A T E3 D2
But E1, E2, E3 can be arranged in
D1 D can be arranged in
The number of arrangements of D1 E1 F E2 A T E3 D2 is
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
three
two
is 8!
3! ways, and
2! ways, so
3! x 2! times

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
Example
How many different permutations can be made using the letters of the words
(i) BOOKS (ii) LOTTO (iii) MATHEMATICS

3. Solution
BOOKS
n = r1 = 2
The number of different permutations is 60.
(ii) LOTTO
n = r1 = r2 = 2
There are 30 different arrangement.

4. 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 = r2 = r3 = 2
There are 90 ways to arrange 2 copies of each of 3 different books on a shelf.

5. Example
How many different 10-letter codes can be made using three a’s, four b’s, and three c’s?
Solution
n = r1 = r2 = r3 = 3
The number of such codes is =
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 =
= ways

6. Example
In how many of the possible permutations of the letters of the
word ADDING are the two D’s:
together,
Separated
Solution
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. = = 5!

7. n (letters) = 12 n(A) = 2, n(R) = 2 n(N) = 2, n(E) = 2
Example
How many different arrangements are there for the letters of
the word ARRANGEMENTS if
a) it begins with “R” and ends with “E”
b) the two letters “E” are separated
c) the two letters “E” and the two letters “A” are together
d) the consonant letters G, M, T, S are together
the two letters “N” occupied both ends
Solution a)
n (letters) = 12
n(A) = 2, n(R) = 2
n(N) = 2, n(E) = 2 R E
1 way
1 way
10 letters
(2! for A, 2! for N)
The NOP =

8. - b) the two letters “E” are separated
The NOP separated = without restriction – E’s together
Without restriction =
E’s together = EE
Thus, the NOP in which the two E’s separated =
(2! for A, 2! for R,
2! for N)
10 letters -

9. c) the two letters “E” and the two letters “A” are togetherEE AA
8 letters
The NOP =
d) the consonant letters G, M, T, S are together
8 letters
GMTS
The NOP =

10. e) the two letters “N” occupied both ends
The NOP =
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?

11. COMBINATION 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).

12. Example 1:
Determine whether each of the following is a permutation or combination:
a) 5 pictures placed in a row.
b) 3 story books picked from a rack.
c) A team of 9 players chosen from a group of 20.
d) The arrangements of the letters in the word OCTOBER.
e) Types of food in a plate taken for lunch consist of rice,
vegetables, chicken curry and prawn paste sambal.
f) Answering questions for Mathematics paper I.
Therefore, we can conclude that permutations are used when order is important and combinations are used when order is not important.
(Permutation)
(Combination)
(Combination)
(Permutation)
(Combination)
(Permutation)

13. Combinations of r object from n objects
3 students could be chosen, in order, from a total of 18 in
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
= 816.
The general result is:
The number of combinations (or selections) of r objects
chosen from n unlike objects is

14. The NOC of 3 objects A, B, C
taken 2 at a time = 3C2 = 3
Consider the following table;
n different object
taken r object
Combination
NOC
Permutation
NOP
A, B 2 AB 1
AB, BA 2
AB, BA, AC,
CA, BC, CB
A, B, C 2
AB, AC, BC 3 6
ABC, ACB,
BCA, BAC,
CAB,CBA
A, B, C 3
ABC 1 6
Thus, we can see that the NOC is always less than the NOP.

15. 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
Therefore the team can be chosen in 1365 ways.

16. 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
Three girls and one boy can be chosen in
Therefore the number of ways of choosing the group if there are more girls than boys is
= 95 ways
or three girls and one boy 6 C
= 15 ways 4 6 4 C x C
= 80 ways 3 1

17. 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 C
Two girls can be chosen in
Two boys can be chosen in
Using the multiplication principle , number of ways of selecting the group
ways 2 7 C
ways 2
= x = 588

18. a) {ab}, {ac}, {ad}, {ae}, {bc}, {bd}, {be}, {cd}, {ce}, {de}
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}

19. 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 =
Number of ways of selecting the defenders =
Number of ways of selecting the midfielders =
Number of ways of selecting the forwards =
Number of ways of selecting the squad
= 2 x 10 x 3 x (by using the principle of multiplication)
= 600

20. Example 7
ABCDEFGH is a regular octagon.
How many triangles can be formed with
the vertices of the octagon as vertices ?
How many diagonals can be drawn by
joining the vertices?
Solution
(a) A triangle is formed by taking 3 of the vertices.
Number of triangles =
(b) A line can be formed by taking any 2 points from the 8
vertices of the octagon.
Number of lines formed =
These 28 lines include the 8 sides of the octagon
Thus the number of diagonals = 28 – 8
= 20 D A B C E F G H

21. 15C7 x 8C5 x 3C3 (2C2 x 13C5 ) x 8C5 x 3C3 = 360360 = 72072 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
15C7
x 8C5
x 3C3 =
a) The NOC =
(2C2 x 13C5 )
x 8C5
x 3C3
= 72072
b) The NOC =

22. 2C1 x 2C1 x 2C1 = 8 x 2C1 x 2C1 x 2C1 = 32 ways 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
the husband and the wife cannot be in the committee at the same time.
Solution
a) The NOC =
4C3 = 4 ways.
4C3
b) The number of selecting 3 groups from 4 groups =
The number of choosing a person from each of the 3
groups selected =
Thus, the NOC =
2C1 x 2C1 x 2C = 8
4C3
x 2C1 x 2C1 x 2C1 = 32 ways

23. = 45 b) The NOC = x = 21 C C C 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
can answer the question if the first 3 questions must be answered.
Solution
a) The NOC = 10
= 45 C 8 3 C 7 C
b) The NOC = x
= 21 5

24. 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 =
= 66 1 5 C 2 5 C 3 5 C 4 5 C 5 C + + + +