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Transcript of NO ONE CAN PREDICT TO WHAT HEIGHTS YOU CAN SOAR EVEN YOU WILL NOT KNOW UNTIL YOU SPREAD YOUR WINGS.
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NO ONE CAN PREDICT TO WHAT HEIGHTS YOU CAN SOAREVEN YOU WILL NOT KNOW UNTIL YOU SPREAD YOUR WINGS
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Chapter 19
PERMUTATIONS AND
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INTRODUCTION
In our daily lives, we often need to enumerate “events” such as the arrangement of objects in a certain way, the partition of things under a certain condition, the distribution of items according to a certain specification and so on.
In this topic, we attempt to formulate a general principle to help us to answer some counting problems like :
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INTRODUCTION
In how many ways can the numbers
0,1,2,3,4,5,6,7,8,9 be used to form a 4-digit number ?
In how many ways can 4 representatives be chosen from a group of 10 students?
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Multiplication Principle
Example 1: There are 3 roads connecting city A and city B, and 4 roads connecting city B and city C. How many ways are there to travel from city A to city C via city B ?
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Multiplication Principle
City C
City B
Total number of ways = 3 x 4 = 12
City A
Solution:
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In this example, we may regard
travelling from A to C via B as a task,
travelling from A to B as stage1, and from B to C as stage 2.
stage1 and stage 2 are necessary or chained/linked for the task to be completed.
Multiplication Principle
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Then we can develop the Multiplication Principle which states that if it requires two necessary stages to complete a task and if there are r ways and s ways to perform stage 1 and stage 2 respectively, then the number of ways to complete the task is r s ways.
Multiplication(r-s) Principle
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If a task/process consists of k linked steps/stages, of which the first can be done in n1 ways, the second in n2 ways,…, the r th step in nr ways, then the whole process can be done in
n1 n2 … nr … n k ways.
Multiplication Principle
Certainly, we can extend the Multiplication Principle to handle a task requiring more than 2 stages:
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If a task can be carried out through n possible distinct(mutually exclusive)
processes(sub-tasks) and if the first process can be done in n1 ways, the second in n2 ways,…, the r th tasks in nr ways, then the task can be done in a total of
n1 + n2 + … + nr + … + nk ways.
- TWO BASIC PRINCIPLES-
Addition Principle
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In how many ways can a man travel from Singapore to Tokyo if there are 4 airlines and 3 shipping lines operating between the 2 cities ?
- TWO BASIC PRINCIPLES-
Addition PrincipleExample 2:
Solution: Task: Travel to Tokyo from Singapore
Task by air – 4 ways
Task by sea – 3 ways
Total number of ways = 4 + 3 = 7
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Note:
Since the man cannot travel by air and
sea at the same time, we say that the two (approaches in carrying out the) tasks are mutually exclusive .
- TWO BASIC PRINCIPLES-
Addition Principle
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In how many ways can we select 2 books of different subjects from among 5 distinct Science books, 3 distinct Mathematics books and 2 distinct Art books?
- TWO BASIC PRINCIPLES-
Addition Principle
Example 2a:
Most questions may be a combination of the two different principles as shown below:
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Solution:There are three cases to consider: Case 1: Select 1 Science & 1 Maths book.
No. of ways = 5 x 3 = 15
Case 2: Select 1 Science & 1 Art book. No. of ways = 5 x 2 = 10
Case 3: Select 1 Maths & 1 Art book. No. of ways = 3 x 2 = 6
Hence, combining Case 1, 2 and 3, total no.of ways of selecting 2 books of different subjects = 15 + 10 + 6 = 31
- TWO BASIC PRINCIPLES-
Addition Principle
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- TWO BASIC PRINCIPLES-
Ex 2a: Solution
ArtScience
ArtMaths
Science
Maths
2 books, 2 different subjects
= 3 x 5 = 15
= 5 x 2 = 10
= 3 x 2 = 6
Total = 31
(A different presentation)
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PERMUTATIONS
Let S be a set of n objects.
A permutation (arrangement) of r objects drawn from the set S( where r < n) is a sub-set of S containing r objects in which the order of the objects in the sub-set is taken into consideration.
DEFINITION
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PERMUTATIONS
Are the following permutations?a) Arranging 3 different books on a shelf.b) Choose 4 books from a list of 10 titles to take away for reading.c) Seating 500 graduates in a row to receive their degree scrolls.d) Forming three-digit numbers using the integers 1, 2, 3, 4, 5.e) Select two students from a CG to attend a talk.
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PERMUTATIONS
Illustration
S = { } = set of 5 objects
Permutations( order within group is considered)
Of 4 objects
different
permutations
of 4objects
Of 3 objects
different
permutations
of 3objects
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- PERMUTATIONS-
Interpretation of
When r objects taken from n different objects are permuted (arranged), the number of possible permutations denoted by nPr
, is given by
nPr = n(n - 1)(n - 2) … (n - r + 1)
=
=
rnP
r)!(n
n!
12...)1)((
12...)1)()(1(...)2)(1(
rnrn
rnrnrnnnn
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- PERMUTATIONS-
Interpretation of rnP
Special case :When all the n different objects are permuted, thenumber of permutations is = = n!
nnP
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- PERMUTATIONS-
Rules Of Permutations
nn
all
different objec
L1: No. of arrangementsof n
in a row / line Ps nt !
nr
r
differ
L2 : No. of a
ent object
rrangementsof out of n
in a row / l nes i P
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In how many ways can the letters A, B, C be arranged ?
Solution:
No. of ways = = 3! = 6
- PERMUTATIONS-
Example 3
33P
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There are 10 vacant seats in a bus. In how many ways can 2 people seat themselves ?
- PERMUTATIONS-
Example 4
Solution:
No. of ways = 10P2
= 10 x 9
= 90
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Find the number of ways of filling 5 spaces by selecting any 5 of 8 different books.
- PERMUTATIONS-
Example 5
Solution:
No. of ways = 8P5
= 8 x 7 x 6 x 5 x 4
= 6720
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How many arrangements of the letters of the word ‘BEGIN’ are there which start with a vowel ?
Solution:The first letter must be either E or I ie 2
ways. The rest can be arranged in 4! ways.
Therefore no. of arrangements = 2 x 4! = 48
- CONDITIONAL PERMUTATIONS-
Example 6
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How many arrangements of the letters of the word ‘BEGIN’ are there which start with a vowel ?
Solution:( A different presentation)
- CONDITIONAL PERMUTATIONS-
Example 6
Arrange using 4 letters ( minus E or I)
E,I
No of arrangements beginning with a vowel
= 2 x 4! = 48
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- PERMUTATIONS-
Rules Of Permutations
nn
all
different objec
L1: No. of arrangementsof n
in a row / line Ps nt !
nr
r
differ
L2 : No. of a
ent object
rrangementsof out of n
in a row / l nes i P
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In how many ways can the letters of the word ‘ORANGE’ be arranged with the restriction that
(i) the 3 vowels must come together (ii) the 3 vowels must not come together (iii) the 3 vowels are all separated?
- CONDITIONAL PERMUTATIONS-
Example 7
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In how many ways can the letters of the word ‘ORANGE’ be arranged with the restriction that (i) the 3 vowels must come together
- CONDITIONAL PERMUTATIONS-
Example 7
Solution:
{ O,A,E } R N G(i)
Strategy: bundle the 3 vowels together as 1 unit but remember you will have to arrange the 3 vowels within this unit in 3! ways.
No. of words in which the 3 vowels are together
= 4! x 3! = 144
Be objective
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In how many ways can the letters of the word ‘ORANGE’ be arranged with the restriction that
(ii) the 3 vowels must not all come together
- CONDITIONAL PERMUTATIONS-
Example 7
Solution:No. of ways to arrange all 6 letters = 6! = 720No. of arrangements in which the vowels are together = 144Therefore, the no. of arrangements in which the vowels are not all together = 720 – 144 = 576
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In how many ways can the letters of the word ‘ORANGE’ be arranged with the restriction that (iii) the 3 vowels are all separated?
- CONDITIONAL PERMUTATIONS-
Example 7
Solution:Qn: How do we separate some objects?Ans: We use the other objects as separators! Thus we use the consonants to separate the 3
vowels as in the configuration below:
C C C
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- CONDITIONAL PERMUTATIONS-
Example 7
C being the position of a consonant and the position of a vowel.
Thus the number of ways of arranging ‘ORANGE’ so
that the 3 vowels are separated = 3! X = 14443P
C C C
The 3 C’s can be arranged in 3! ways; and we can use 4 positions to put in the 3 vowels in ways.4
3P
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Given six digits 1,2,4,5,6,7. Find the number of 6-digit numbers which can be formed using all 6 digits without repetition if
i) there is no restriction; ii) the numbers are all even; iii) the numbers begins with ‘1’ and end
with ‘5’.
- CONDITIONAL PERMUTATIONS-
Example 8
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Solution:(i) No. of ways = 6! = 720
- CONDITIONAL PERMUTATIONS-
Example 8
(ii) For each no. formed to be even, the last digit must be occupied by 2, 4 or 6 i.e. there are 3 ways to fill the last digit.
The first 5 digits can be filled in 5! ways. Therefore, the total no. of even nos. than
can be formed = 3 x 5! = 360
2,4,6
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(iii) The first digit(1) and the last digit(5) are fixed (no arrangement needed). The remaining 4 digits can be filled in 4! ways. Therefore, the total no. of such nos. formed = 1 x 4! = 24
- CONDITIONAL PERMUTATIONS-
Example 8
1 5
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COMBINATIONS
A Combination (Selection) of a given number of articles is a sub-set of articles selected from those given where the order of the articles in the subset is not taken into consideration.
Examples of combinations a) Pick a set of three integers from the numbers 1, 2, 3, 4, 5. b) Choose a committee of 3 persons from a group of 10 people. c) From a box of 200 lucky draw tickets, 3 to be drawn as consolation prize.
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COMBINATIONS
Consider the number of permutations of three different books A, B, C on a self:
ABC, ACB, BAC, BCA, CAB, CBA ----- 3! = 6 ways If order is not important, then there is only one way to
choose all the three different books,
i.e. simply {A, B, C}
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COMBINATIONS
Possible selections or combinations of 3 objects drawn from the set { A, B, C, D } are
43C
Illustration
{ A, B, C } , { A, B, D }, { A, C , D},{ B, C , D}.
No. of possible selections or combinations of 3 objects drawn from the set { A, B, C, D }
= 4 = = 4P3 / 3!
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COMBINATIONS
Possible selections or combinations of 2 objects drawn from the set { A, B, C, D } are { A, B } , { A, C }, { A, D } , { B, C }, { B, D } , { C, D }.
= 6 or = 4P2 / 2! 4
2C
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COMBINATIONS
Possible selections or combinations of 2 objects drawn from the set { A, B, C, D,E } are
52C
An exercise
{A, B} , {A, C} , {A, D} , {A, E} , {B, C} ,
{B, D} , {B, E} , {C, D} , {C, E} , {D, E}
No. of possible selections or combinations of 2 objects drawn from the set {A, B, C, D, E }
= 10 = = 5P2 / 2!
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COMBINATIONS
Possible selections or combinations of 3 objects drawn from the set { A, B, C, D,E } are
53C
An exercise
{A, B, C} , {A, B, D} , {A, B, E} , {A, C, D} {A, C, E} , {A, D, E} , {B, C, D} , {B, C, E} {B, D, E} , {C, D, E}
No. of possible selections or combinations of 2 objects drawn from the set {A, B, C, D, E }
= 10 = = 5P3 / 3!
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COMBINATIONS
Possible selections or combinations of 6 nos. drawn from the set { 1,2,3,4,…,45 } are
456C
Illustration
impossible to list out without the aid of a computer.
No. of possible selections or combinations of 6 nos. drawn from the numbers 1 to 45 inclusive = = 8 145 060
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COMBINATIONS
The number of combinations of r objects taken from n unlike objects is , where
rnC
!
)1)...(2)(1(
)!(!
!
r
rnnnn
rnr
n
!r
Prn
rnC =
=
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In how many ways can a committee of 3 be chosen form 10 persons ?
Solution:
No. of ways = 10C3 = 120
- COMBINATIONS-
Example 9
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In how many ways can 10 boys be divided into 2 groups (i) of 6 and 4 boys (unequal sizes)
Solution:
- COMBINATIONS-
Example 10
Select
First group of 6
Select
Second group of 4
No. of ways of forming groups = 10C6 X 4C4
= 210
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In how many ways can 10 boys be divided into 2 groups (ii) of 5 and 5 boys (equal sizes)
Solution:
- COMBINATIONS-
Example 10
Select
First group of 5
Select
Second group of 5
No. of ways of forming groups = (10C5 X 5C5)/2! = 126
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A committee of 5 members is to be selected from 6 seniors and 4 juniors. Find the number of ways in which this can be done if a) there are no restrictions,b) the committee has exactly 3 seniors,c) the committee has at least 1 junior
- COMBINATIONS-
Example 11
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- COMBINATIONS-
Example 11
Solution:
a) If there are no restrictions, the 5 members can be selected from the 10 people in 10C5 = 252 ways
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- COMBINATIONS-
Example 11
Solution:b) If the committee has exactly 3 seniors,
these seniors can be selected in 6C3 ways.The remaining 2 members must then be juniors which can be selected from the 4 juniors in 4C2 ways.By the multiplication principle, the no. of committees with exactly 3 seniors is
6C3 4C2 = 120
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- COMBINATIONS-
Example 12
Solution:c) No. of committees with no juniors
= 4C0 6C5 = 6No. of committees with at least 1 junior = (total no. of committees) – (no. of committees with no junior)= 252 – 6 = 246
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