Permutations where some objects are repeatedPermutations where some objects are repeated

Consider the number of permutations of the letter

D E F E A T E DD E F E A T E D

There are E’s and D’s.

The number of permutations of DD1 1 E E1 1 F E F E2 2 A T E A T E3 3 D D22

But EE11, E, E22, E, E33 can be arranged in DD11 D D22 can be arranged in

The number of arrangements of DD11 E E11 F E F E22 A T E A T E33 D D22 is the number of arrangements of D E F E A T E DD E F E A T E D. Therefore the number of permutations of D E F E A T E D D E F E A T E D is

3360!2!3

!8

threethree twotwo

is 8!8!

3!3! ways, and

2!2! ways, so

3! x 2! times3! x 2! times

Generalising this argument we see that, The number of permutations of n objects comprising of

rr1 1 identical objects, rr2 2 identical objects, ..……….,

rrkk identical objects is

!..........!!

!

21 krrr

n

Example Example

How many different permutations can be made using the How many different permutations can be made using the letters of the wordsletters of the words

(i) BOOKS (ii) LOTTO (iii) MATHEMATICS(i) BOOKS (ii) LOTTO (iii) MATHEMATICS

Solution

(i) BOOKS

n = 5 r1 = 2

The number of different permutations is 60.

(ii) LOTTO

n = 5 r1 = 2 r2 = 2

There are 30 different arrangement.

60)3)(4)(5(!2

!5

!

!

r

n

30

2

345

!2!2

!5

!!

!

21

rr

n

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 r3 = 2

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

90

22

3456

!2!2!2

!6

!!!

!

321

rrr

n

Example Example How many different 10-letter codes can be made using three a’s, four b’s, and three c’s?

SolutionSolution

n = 10 n = 10 rr11 = 3 = 3 rr22 = 4 = 4 rr33 = 3 = 3

The number of such codes is

= 4200.

2323

5678910

!3!4!3

!10

Example:Example:In how many ways can 3 red, 4 blue and 2 green pens be In how many ways can 3 red, 4 blue and 2 green pens be distributed among nine students seated in a row if each distributed among nine students seated in a row if each student receives one pen?student receives one pen?

SolutionSolutionThe NOP =The NOP = !2!4!3

!9= 1260 ways

Example Example In how many of the possible permutations of the letters of the word ADDING are the two D’s:(i) together, (ii) Separated

SolutionSolution

(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.

= -

= 240.

!2

!65!

ExampleExampleHow many different arrangements are there for the letters of the word ARRANGEMENTS ARRANGEMENTS if

a) it begins with “R” and ends with “E”b) the two letters “E” are separatedc) the two letters “E” and the two letters “A” are togetherd) the consonant letters G, M, T, S are togethere) the two letters “N” occupied both ends

SolutionSolution

a) a) RR EE

10 letters

The NOP = The NOP = 1!2!2

!101

1 way1 way1 way1 way

(2! for A, 2! for N)(2! for A, 2! for N)

n (letters) = 12n(A) = 2, n(R) = 2n(N) = 2, n(E) = 2

b) the two letters “E” are separated

The NOP separated = without restriction – E’s togetherThe NOP separated = without restriction – E’s together

Without restriction = Without restriction =

E’s togetherE’s together = =

EEEE

Thus, the NOP in which the two E’s separatedThus, the NOP in which the two E’s separated

= =

!2!2!2!2

!12

10 letters!2!2!2

!11 (2! for A, 2! for R, (2! for A, 2! for R, 2! for N)2! for N)

!2!2!2!2

!12

!2!2!2

!11-

c) the two letters “E” and the two letters “A” are together

EE AA 8 letters

The NOP = 11!2!2

!10

d) the consonant letters G, M, T, S are together

GMTS8 letters

The NOP =!2!2!2!2

!9 !4

e) the two letters “N” occupied both ends

NN NN

10 letters

The NOP =!2!2!2

!10 11

ExerciseExercise

A dancing contest has 11 competitors, of whom three are A dancing contest has 11 competitors, of whom three are Americans, two are Malaysians, three are Indonesians, and Americans, two are Malaysians, three are Indonesians, and three are Italians. If the contest result lists only the three are Italians. If the contest result lists only the nationality of the dancers, how many outcomes are possible?nationality of the dancers, how many outcomes are possible?

Combinations of a set of objectsCombinations of a set of objects

• A combination is a selectionselection 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, RQPPQR, 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: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)(Permutation)

(Permutation)(Permutation)

(Permutation)(Permutation)

(Combination)(Combination)

(Combination)(Combination)

(Combination)(Combination)

3 students could be chosen, in order, from a total of 18 in .!15

!18ways

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

!3!15

!18= 816.

The general result is:

The number of combinations (or selections) of r objects The number of combinations (or selections) of r objects chosen from n unlike objects ischosen from n unlike objects is

!!

!

rrn

nC r

n

Combinations of r object from n objects Combinations of r object from n objects

n different object

taken r object

Combination NOC Permutation NOP

Consider the following table;

!!

!

rrn

nC r

n

!!

rn

nP r

n

A, B

A, B, C

A, B, C

2

3

2

AB

ABC

AB, AC, BC

1

3

1 AB, BA

AB, BA, AC, CA, BC, CB

ABC, ACB, BCA, BAC,CAB,CBA

6

6

2

Thus, we can see that the NOC is always less than the NOP.Thus, we can see that the NOC is always less than the NOP.

The NOC of 3 objects A, B, The NOC of 3 objects A, B, CC

taken 2 at a time = taken 2 at a time = 33CC22 = 3 = 3

Example 2Example 2

A quiz team of four is chosen from a group of 15 students. In how many ways could the team be chosen?

SolutionSolution

Therefore the team can be chosen in 1365 ways.

!4!11

!1515

C 4

1365

Example 4Example 4A 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?

SolutionSolution

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

15 + 80 = 95 ways

or three girls and one boy

6C 4

= 15 ways15 ways

36C x

4C1

= 80 ways80 ways

Example 3Example 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?

SolutionSolution

Two girls can be chosen in

Two boys can be chosen in

Using the multiplication principle , number of ways of selecting the group

8C

2

7C

2

ways

ways

= 28 x 21 = 588

27

28 CC

!2!5

!7

!2!6

!8

Example 5Example 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 6Example 6

In a football training squad of 24 people, 3 are goalkeepers, 7 are 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 defenders, 6 are midfielders and 8 are forwards. A final squad of 16 selected for a match must consist of 2 goalkeepers, 4 16 selected for a match must consist of 2 goalkeepers, 4 defenders, 5 midfielders and 5 forwards. Find the number of defenders, 5 midfielders and 5 forwards. Find the number of possible selections if one particular goalkeeper, 2 particular possible selections if one particular goalkeeper, 2 particular defenders, 3 particular midfielders and 3 particular forwards are defenders, 3 particular midfielders and 3 particular forwards are automatically selected.automatically selected.

SolutionSolutionNumber 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 squadNumber of ways of selecting the squad = 2 x 10 x 3 x 10 = 2 x 10 x 3 x 10 (by using the principle of multiplication)(by using the principle of multiplication) = 600= 600

212

11 CC

1025

22 CC

323

33 CC

1025

33 CC

Example 7Example 7

ABCDEFGH ABCDEFGH is a regular octagon. is a regular octagon.(a)(a) How many triangles can be formed with How many triangles can be formed with

the vertices of the octagon as vertices ?the vertices of the octagon as vertices ?(b)(b) How many diagonals can be drawn by How many diagonals can be drawn by

joining the vertices? joining the vertices?

SolutionSolution (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

5638 C

2828 C

D

AB

C

EF

G

H

Example 8Example 8

15 students are divided into 3 groups, with A having 7 15 students are divided into 3 groups, with A having 7 students, group B having 5 students and group C having 3 students, group B having 5 students and group C having 3 students. Find the number of ways to formstudents. Find the number of ways to forma) the 3 groupsa) the 3 groupsb) the 3 groups with 2 given students must be in group A.b) the 3 groups with 2 given students must be in group A.

SolutionSolution

a) The NOC =

b) The NOC =

15C7 = 360360

(2C2 x 13C5 ) = 72072

x 8C5 x 3C3

x 8C5 x 3C3

Example 9Example 9A 3 member committee is to be formed from 4 couples of A 3 member committee is to be formed from 4 couples of husband and wife. Find the possible number of committees husband and wife. Find the possible number of committees that can be formed ifthat can be formed if

a) all the members are mena) all the members are menb)b) the husband and the wife cannot be in the committee at the husband and the wife cannot be in the committee at

the same time.the same time.

SolutionSolution

a) The NOC =

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 =

4C3 = 4 ways.

4C3

2C1 x 2C1 x 2C1 = 8

x 2C1 x 2C1 x 2C1 = 32 ways 4C3

Example 10Example 10

In a test, a candidate is required to answer 8 out of 10 In a test, a candidate is required to answer 8 out of 10 questions. Find the number of ways a candidatesquestions. Find the number of ways a candidates

a) can answer the questionsa) can answer the questionsb)b) can answer the question if the first 3 questions must be can answer the question if the first 3 questions must be

answered.answered.

SolutionSolution

a) The NOC = a) The NOC = 8

10C

5

7C

= 45= 45

b) The NOC =b) The NOC = = 21= 213

3C xx

Example 11Example 11

In how many ways can a teacher choose one or more In how many ways can a teacher choose one or more students as a prefects from 5 eligible students?students as a prefects from 5 eligible students?

SolutionSolution

The NOC is may be one or two or three or four or five person The NOC is may be one or two or three or four or five person chosenchosen

==

== 66 66

1

5C 2

5C 5

5C4

5C3

5C+ + + +