Elementary Mathematics - distance-time graphs & speed-time graphs

Below are the various types of distance-time graphs and its corresponding speed-time graphs.
Constant speed
At rest
Accleration at a constant rate
Deceleration at a constant rate
Deceleration at varying rate

Additional Mathematics - Inverse function

The inverse function is a one-to-one function. Each object in the domain is map to each object in the range. To satisfy this requirement, the function f(x), should not have any turning point. That is, the gradient cannot be equal to zero. This can be verified by computing the the differentiated results - dy/dx) should not be equal to zero.

Eg one-to-one function
Domain---------Range
1---------------3
2---------------5
3---------------7
4---------------9

Eg not one-to-one function
Domain----------Range
1
2---------------5
3---------------7
4---------------9

You can imagine the Range to be the mirror image of the Domain. The mirror is the line
y = x.

Inverse function of a linear equation
Eg, f(x) = x - 5
y = x - 5
exchange x to y and y to x,
x = y - 5
rearrange the equation,
x + 5 = y
y = x + 5
Inverse function of a quadratic equation
Eg, f(x) = x2 + 2x + 5
y = x2 + 2x + 5
y = (x + 1)2 + 4
exchange x to y and y to x
x = (y + 1)2 + 4
rearrange the equation
x - 4 = (y + 1)2
sq root(x - 4) = y + 1
[sq root(x - 4)] -1 = y
y = [sq root(x - 4)] - 1
Inverse function of a cubic equation
Eg f(x) = x3 + x - 1
y = x3 + x - 1
exchange x to y and y to x
x = y3 + y - 1
For cubic equation, it is impossible to re-arrange the equation to isolate y on the left hand side and the various x on the right hand side.
f-1(9), 9 = y3 + y - 1
using trial and error method, determine the value of y.
if y = 1, 1 + 1 - 1 not equal to 9
if y = 2, 8 + 2 - 1 = 9
Therefore, f-1(9) = 2

Oops! sorry for the untidy graphs!

Additional Mathematics - Integration

Integration is the reverse process of differentiation.
For eg,
when we differentiate 'A', we get 'B'.
when we integrate 'B', we get 'A'

To solve integration problems, we need a copy of the formulaes with us. I find the integration formulae list in the Longman A Maths guide comprehensive.

But there are some integration questions that I am unable to solve, even with the formulae.
For eg, Integrate (x)/[sq root(x - 1)].
The answer is y = (x + 2)[sq root(x - 1)]
I know the answer because this is a guided question.
Part 1 of the question requires me to differentiate y = (x + 2)[sq root(x - 1)]. The answer is (3/2)(x)/[sq root(x - 1)]
And Part 2 of the question requires me to integrate (x)/[sq root(x - 1)], which is the original function y.

Thankfully, I have not come across a question in the past 10 years of Additional Mathematics GCE 'O' level examination that requires integration of a complicated function without providing any guides. ^o^

In the integration formulaes provided, integrate (1/x) is (ln x + C). But, it does not say what is the answer for integrating (ln x). Luckily, I found an answer through a guided question. And, I thought I should record in this space for my readers.

Integrate (ln x) = (x)(ln x) - x + C
Integrate (x)(ln x) = (1/2)(x2)(ln x) - (1/4)(x2) + C
Integrate (x2)(ln x) = (1/3)(x3)(ln x) - (1/9)(x3) + C
follow the pattern.

Happy Integration ^o^

Poems - Jul 2009

现今社会追名利，我的偶像是明义。明义毕业莱佛仕,出家修行在佛寺.掌管仁慈帮病人,电视筹款当超人.气车洋房会员卡,宠物养匹澳洲马.还好养马非马子,难到最爱是鸭子? 为了五万帮志恒,如今法庭被质问.杜莱事件刚刚过,国人又气又难过. 人非圣贤谁无过, 认错最重要改过.

My friend's poem collection

I have a friend. He is like the ancient poet whom I have watched on television drama when I was a little girl. He sings a poem over everything he comes across. This space is dedicated to this friend - My friend's poem collection.

Preschool - Mathematics

In mainstream education, the Ministry of Education detailed the syllabus for a child's learning. For preschoolers, the MCYS do not provide the answers on what preschoolers should learn. The preschools are free to provide their own syllabus.

I have done a comparison of the syllabus of a few preschools, references and primary school textbooks and this is my list of 'must learn' for Primary 1 readiness on Mathematics.

1. Counting 1 to 150
2. Spelling the numbers in words 1 to 100
3. Patterns
4. Addition
5. Subtraction
6. Multiplication 1 to 50
7. Clock
8. Dollars and Cents

I am in favour of the Kumon workbook, available in Popular bookstore at $10.90 each. I like their techniques on imparting the Mathematics skills to preschoolers. The workbooks are also printed on quality papers and the layouts are attactive. However, I find the workbooks voluminous for a young child. There are many repetition to allow sufficient practices of each topic. My child takes advantage of these repetition and choose to copy the answers instead. So, I used the workbooks selectively to sustain a child's interest on the subject.

Additional Mathematics - Combinations and Permutations (no repetition)

Questions on Combinations and Permutations gives the stories and requires you to provide the solutions. Comparison and studying the reverse gives better understanding of combinations and Permutations. Read on. It helps.

Eg A, B, C
1. Combination: 3C2 = 3 = AB, AC, BC
The order is not important. It does not matter who comes first. The order is interchangeable ie AB = BA, AC = CA, BC = CB

2. Permutation: 3P2 = 6 = AB, BA, AC, CA, BC, CB
The order is important. It shows who comes first and who is next.

3. Factorial:
2! = 2P2 = 2 = AB, BA
3! = 3P3 = 6 = ABC, ACB, BAC, BCA, CAB, CBA
The order is important.

(1), (2) and (3) above looks at 1 group of object only. They look simple. Complication comes in when there are 2 or more groups of objects. When this happens, we need to use a mixture of the above 3 rules. The following scenarios are important as they are the trends of GCE 'O' Level examination questions.

4. Multiplication of Permutation and Factorial
Group 1: A, B, C
Group 2: 1

3P2(Group 1) = 6 = AB, AC, BC, BA, CA, CB
2! (both groups) = 2 = G1G2 or G2G1
3P2(Group 1) x 2!(both groups) = 6 x 2 = 12
= AB1, 1AB, AC1, 1AC, BC1, 1BC, BA1, 1BA, CA1, 1CA, CB1, 1CB

5. Multiplication of Permutation with Permutation and Factorial
Group 1: A, B, C
Group 2: 1, 2, 3, 4

3P2(Group 1) = 6 = AB, AC, BC, BA, CA, CB
4P2(Group 2) = 12 = 12, 21, 13, 31, 14, 41, 23, 32, 24, 42, 34, 43
2!(Group 1 & Group 2) = 2 = G1G2, G2G1

3P2(Group 1) x 4P2(Group 2) x 2!(Group 1 & Group 2) = 6 x 12 x 2 = 144

AB12 12AB AB21 21AB AB13 31AB AB14 14AB AB41 AB41 AB23 23BA AB32 32AB AB24 24AB
AC12 12AC AC21 21AC AC13 31AC AC14 14AC AC41 AC41 AC23 23AC AC32 32AC AC24 24AC
BC12 12BC BC21 21BC BC13 31BC BC14 14CB BC41 41BC BC23 23BC BC32 32BC BC24 24BC
BA12 12BA BA21 21BA BA13 31BA BA14 14AB BA41 41BA BA23 23BA BA32 32BA BA24 24BA
CA12 12CA CA21 21CA CA13 31CA CA14 14CA CA41 41CA CA23 23CA CA32 32CA CA24 24CA
CB12 12CB CB21 21CB CB13 31CB CB14 14CB CB41 41CB CB23 23CB CB32 32CB CB24 24CB

AB42 42AB AB34 34AB AB43 43AB AB31 31AB
AC42 42AC AC34 34AC AC43 43AC AC31 31AC
BC42 42BC BC34 34BC BC43 43BC BC31 31BC
BA42 42BA BA34 34BA BA43 43BA BA31 31BA
CA42 42CA AC34 34AC AC43 43AC AC31 31AC
CB42 42CB CB34 34CB CB43 43CB CB31 31CB Phew!

Now that you have mastered the patterns and properties of Combination, Permutation and Factorial and their multiplication. The challenges lies with the identification on what are the groups. Practice makes perfect.

If you like the above or you have any comments, pls let me know at SwallowTuition@gmail.com. I would like to hear from you. Thank You.

Preschool - Phonics

During my childhood, I was not exposed to phonics. Today, phonics is a popular topic for pre-schoolers. The lower primary schools textbooks cover the advanced level of phonics too.

So, what is phonics?
Phonics is a step by step guide to help preschoolers read and spell.

My child's pre-school introduced phonics at the age of 4. Her teacher was always complaining to me that she was unable to remember the phonics sound. I tried to coach her myself, and I faced the same problem as the pre-school teacher. One day, when I have a chance to talk to some parents of my child's classmates, I realised that my child was not alone.

I searched for books and internet to learn ways and means to help my child. I have a good friend who helped me on this too. I took more than a year to master the correct techniques on teaching phonics.

Looking back, I tried to analyse my child's problems. Firstly, she may not be ready to learn phonics at 4 year-old. All of us are good at something and poor at others, we are simply not perfect. Also, the child is not to blame for her weaknesses, the teacher played a part too. Her teacher was unable to develop her interest in the subject and helping her learn. Just like adults, when we have get over the basic, we will be motivated to go deeper into the subject.

Additional Mathematics - Combinations and Permutations (Examples on Permutations)

Permutations - order does matter
You can recognise such question if the question mention some orders in the arrangement of the objects. So, you can use the 'permutation' formulae. Noticed that I have a systematic approach of solving my permutation questions. Try it, it is quite useful.

Listed below are solutions to GCE 'O' level past 10 years examination questions on permutations.

Eg 1
In how many ways can 6 different Chinese books, 3 different English books and 4 different Malay books be arranged on a shelf if all the books in the same language come together?

The 6 Chinese books can be arranged in 6! ways.
The 3 English books can be arranged in 3! ways.
The 4 Malay books can be arranged in 4! ways.
The books of 2 different languages can be arranged in 3! ways.
Number of ways the books can be arranged on the shelf = 6! x 3! x 4! x 3! = 622,080

Eg 2
In how many ways can 4 boys and 4 girls be arranged in a line so that the boys and the girls are placed alternatively?

B G B G B G B G
G B G B G B G B

Each group of 4 can be arranged in 4! ways: 4! x 4! = 576 ways.
As there are 2 combinations, there are 2! ways to arrange the children.
So, the total number of ways the children can be arranged is : 576 x 2! = 1,152.

Eg 3
Find the number of distinct arrangements of all the letters of the word MONDAY. Find the number of these arrangements in which
a) the letters M and Y are together.
b) the letters M and Y are not together.

a)Group 1 : M, Y
Group 2 : 'MY', O, N, D, A

2!(Group 1) = 2 = MY, YM
5!(Group 2) = 120

So, the number of arrangements = 120 x 2 = 240

b) when there is no restriction : 6! = 720
720 - 240 = 480

Eg 4
John would like to arrange 2 Physics textbooks, 3 Economics textbooks and 4 Chemistry textbooks on a newly-purchased bookshelf. Find the number of ways in which the textbooks can be arranged if
a) the Economics textbooks must be next to each other,
b) the Physics textbooks must be separated from each other,
c) textbooks of the same subject must be next to each other,
d) the Economics textbooks must be separated from each other.

Group 1: 2 Physics textbooks
Group 2: 3 Economics textbooks
Group 3: 4 Chemistry textbooks

a)
3!(Economics textbooks) = 6
7!(Economics group + 6 other textbooks) = 5,040
5,040 x 6 = 30,240

b) Physics textbook together = 2! = 2
Physics textbook together + 7 other books = 8! = 40,320
Total arrangements = 2 x 40,320 = 40,320 x 2 = 80,640

no restriction = 9! = 362,880
362,880 - 80,640 = 282,240

c) Physics : 2! = 2
Economics : 3! = 6
Chemistry : 4! = 24
3 subjects : 3! = 6
2 x 6 x 24 x 6 = 1,728

d) 2 Economics textbook together : 3P2 = 6
2 Econ and others : 8! = 40,320
40,320 x 6 = 241,920

362,880 - 241,920 = 120,960

Eg 5 (N02/P2/5b)
Find how many different odd 4-digit numbers less than 4000 can be made from the digits 1,2,3,4,5,6,7 if no digit may be repeated.

1st digit : 1, 2 or 3
2nd digit : any odd or even number (less 1st & 4th digit)
3rd digit : any odd or even number (less 1st, 2nd & 4th digit)
4th digit : any odd number (less 1st digit)

1000-1999 : 1 x 5 x 4 x 3 = 60
2000-2999 : 1 x 5 x 4 x 4 = 80
3000-3999 : 1 x 5 x 4 x 3 = 60
Total = 200

Eg 6 (N99/P2/24c (part 2))
An art gallery has 9 paintings by a famous artist. A selection of 4 of these are to be shown in an exhibition. These 4 paintings, including a special one, are exhibited in a line. Find the number of arrangements of these 4 paintings in which the special painting does not come at either end.

The arrangements are:
a) N S N N, or
b) N N S N

8P3(non-special paintings) = 336
As there are 2 similar combination, there are 2! ways to arrange paintings
So, the total number of ways to arrange the books are : 336 x 2 = 672

Eg 7 (N2004/P1/7a)
Find the number of different arrangements of the 9 letters of the word SINGAPORE in which S does not occur as the first letter.

There are 8 ways to arrange S = 1 x 8 = 8
There are 8 ways to arrange 8 alphabets(INGAPORE) : 8! = 40,320

So, the total number of ways is 40,320 x 8 = 322,560

Eg 8 (N2005/P2/11b)
Calculate the number of different 5-digit numbers which can be formed using the digits 0,1,2,3,4 without repetition and assuming that a number cannot begin with 0. How many of these 5-digit numbers are even?

There are 4 ways to arrange 0 : 4 x 1 = 4
There are 4 permutations to arrange 4 numbers (1,2,3,4) : 4! = 24
So, the total number of ways is 24 x 4 = 96

How many of these 5 digits are even?

There are 2 groups:
a) even group(2,4) : E
b) others(0,1,3) : 3

The combinations are:
EEOOO OEEOO OOEEO OOEOE
EOEOO OEOEO OOOEE
EOOEO OEOOE
EOOOE

There are 3 other numbers (0,1,3) : 3
There are 2 ways to arrange 2 even numbers (2,4) : 2!
There are 10 ways to arrange the numbers : 3 x 2 x 10 = 60

Eg 9 (N2007/P2/8aii)
Each of seven cards has on it one of the digits 1,2,3,4,5,6,7; no two cards have the same digit. Four of the cards are selected and arranged to form a 4-digit number.
(i) How many different 4-digit numbers can be formed in this way?
(ii) How many of these 4-digit numbers begin and end with an even digit?

(i) 7P4 = 840

2 Groups:
Group 1 : even numbers (2,4,6)
Group 2 : odd number + 1 of group 1 {1,3,5,7 +(2,4,6)}

(ii)compute ways of AEEA
If 2 even numbers are together : 3P2 = 6
2+1=3 blocks for balance 6 numbers : 6P3 = 120
120 x 6 = 720
If there is no restriction: 840-720 = 120.
But, there is a possibility that the E is not in the centre but at the side ie
AAE, or
EAA
So, need to eliminate the possibility that the 2 AA are together.
6P2 = 30
As there are 2 combinations, ie AAE or EAA, 30 x 2 = 60
So, 120 - 60 = 60

Additional Mathematics - Combinations and Permutations (Examples on Combination)

Combination - Order does not matter
You can recognise if a question is a combination question if it specifically says 'order does not matter' or the question did not tell you in what order should you arrange the objects.

Listed below are solutions to GCE 'O' level past 10 years examination questions on combinations.

Eg 1,
A committee of 6 is chosen from 10 men & 7 women so as to contain at least 3 men & 2 women. How many ways can this be done.

The combinations are:
1) 4 men and 2 women, or
2) 3 men and 3 women

Under the probability rule, and is multiplication(x) and or is addition(+).
a) 4 men and 2 woman : 10C4 x 7C2 = 4,410
b) 3 men and 3 woman : 10C3 x 7C3 = 4,200
4,410 + 4,200 = 8,610

Eg 2,
There are 4 pigeon holes marked W, X, Y and Z. In how many ways can 14 different books be arranged so that 5 of them are in W, 4 of them are in X, 3 of them are in Y and the rest are in Z? The order of books in each hole does not matter.

hole W : 5 books, and
hole X : 4 books, and
hole Y : 3 books, and
hole Z : 2 books.
Total : 14 books

hole W(choose from 14 books) : 14C5 = 2,002 ways
hole X(14-5=9 books left to choose) : 9C4 = 126 ways
hole Y(9-4=5 books left to choose) : 5C3 = 10 ways
hole Z(5-3=2 books, nothing to choose) : 1

So, the total number of ways of arranging the books = 2,002 x 126 x 10 x 1
= 2,522,520

Eg 3
A committee of 7 members is to be selected from 8 men and 5 women. Calculate the number of ways in which this can be done if
a) there are no restrictions,
b) the committee is to have exactly 4 men,
c) the committee is to have at least 1 women,
d) the committee is to have more women than men.

a) 13C7 = 1,716
b) 8C4 x 5C3 = 700
c) All men committee : 8C7 = 8
1,716-8 = 1,708
d) The combinations are:
i) 3 men 4 women = 8C3 x 5C4 = 280
ii) 2 men 5 women = 8C2 x 5C5 = 28
Total : 280 + 28 = 308

Eg 4
A Club formed by some undergraduates intends to send a group of 10 members to attend an exhibition. The club has a total of 20 members, of which 8 are Arts undergraduates, 7 are Business undergraduates and 5 are Engineering undergraduates. Find the number of ways to choose the group if
a) the selection process is random,
b) 5 Arts, 3 Business and 2 Engineering undergraduates must be chosen
c) an equal number of Business and Engineering undergraduates must be chosen,
If the club intends to send a group comprising 5 Arts, 3 Business and 2 Engineering undergraduates, calculate the number of ways in which
d) the group may be chosen if it is to include 3 specified Arts and 1 specified Business undergraduates,
e) the 5 Arts undergraduates may be chosen from the 8 available, given that 2 particular Arts undergraduates cannot be chosen together.

20 memebers: 8 Arts + 7 Biz + 5 Eng
a) 20C10 = 184,756
b) 8C5 x 7C3 x 5C2 = 19,600
c) The combinations are:
8A+1B+1E, or
6A+2B+2E, or
4A+3B+3C, or
2A+4B+4C, or
0A+5B+5C

8C8 x 7C1 x 5C1 = 35
8C6 x 7C2 x 5C2 = 5880
8C4 x 7C3 x 5C3 = 24,500
8C2 x 7C4 x 5C4 = 4,900
7C5 x 5C5 = 21
Total = 35,315

d) 2A and 3A(specified) and 2B and 1B(specified) and 2E
5C2 x 6C2 x 5C2 = 1,500
e) no restriction in arts students : 8C5 = 56
if the 2 arts students were chosen together : 6C3 = 20
if the 2 arts students were not chosen together : 56 - 20 = 36
3B and 2E = 7C3 x 5C2 = 350
350 x 36 = 12,600

Eg 5 (N99/P2/24c)
An art gallery has 9 paintings by a famous artist. A selection of 4 of these are to be shown in an exhibition. Calculate the number of different selections that can be taken if (i) there are no restrictions, (ii) one special painting must be included.

(i) 9C4 = 126
(ii) 1 special painting and (choose the other 3 painting from the balance 8 paintings)
1 special painting = 1
choose 3 paintings from 8 paintings : 8C3 = 56
So, the number of selections are : 1 x 56 = 56

Eg 6 (N2002/P2/5a)
The producer of a play requires a total cast of 5, of which 3 are actors and 2 are actresses. He auditions 5 actors and 4 actresses for the cast. Find the total number of ways in which the cast can be obtained.

actors : 5C3 = 10, and
actresess : 4C2 = 6
So, the total number of ways the cast can be obtained are : 10 x 6 = 60

Eg 7 (N2003/92/8)
A garden centre sells 10 different varieties of rose bush. A gardener wishes to buy 6 rose bushes, all of different varieties.
(i) Calculate the number of ways she can make her selection.
Of the 10 varieties, 3 are pink, 5 are red and 2 are yellow. Calculate the number of ways in which her selection of 6 rose bushes could contain
(ii) no pink rose bush
(iii) at least one rose bush of each colour.

(i) 10C6 = 210
(ii) 10 rose : 3 pink + 5 red + 2 yellow
no pink rose, so left with 7 roses to choose for 6 rose : 7C6 = 7
(iii)
a) P P P R R Y, or
b) P P R R R Y, or
c) P R R R R Y, or
d) P R R R Y Y, or
e) P P R R Y Y, or
f) P P P R Y Y, or

a) 3C3 x 5C2 x 2C1 = 20
b) 3C2 x 5C3 x 2C1 = 60
c) 3C1 x 5C4 x 2C1 = 30
d) 3C1 x 5C3 x 2C2 = 30
e) 3C2 x 5C2 x 2C2 = 30
f) 3C3 x 5C1 x 2C2 = 5
Total = 175 ways

Eg 8 (N2004/P1/7b)
3 students are selected to form a chess team from a group of 5 girls and 3 boys. Find the number of possible teams that can be selected in which there are more girls than boys.

a) G G B, or
b) G G G

a) GGB : 5C2 x 3C1 = 30
b) GGG : 5C3 = 10
So the number of teams are 30 + 10 = 40

Eg 9 (N2005/P2/11a)
Each day a newsagent sells copies of 10 different newspapers, one of which is The Times. A customer buys 3 different newspapers. Calculate the number of ways the customer can select his newspapers. (i) if there is no restriction (ii) if 1 of the 3 newspapers is The Times.

(i) 10C3 = 120
(ii) The Times & choose 2 more other newspapers from the other 9 newspapers
The Times = 1
Choose 2 newspapers from 9: 9C2 = 36

So, the number of ways to select the newspapers are : 1 x 36 = 36 ways.

Eg 10 (N2006/P2/10)
a) How many different four-digit numbers can be formed from the digits 1,2,3,4,5,6,7,8,9 if no digit may be repeated?
b)In a group of 13 entertainers, 8 are singers and 5 are comedians. A concert is to be given by 5 of these entertainers. In the concert there must be at least 1 comedian and there must be more singers than comedians. Find the number of different ways that the 5 entertainers can be selected.

a) 9C1 x 8C1 x 7C1 x 6C1 = 3,024
b) 13 entertainers : 8 singers + 5 comedians
concert (5 entertainers) : >= 1 comedian, singers > comedians

The combinations are:
a) C S S S S, or
b) C C S S S

a) 5C1 and 8C4 = 350
b) 5C2 and 8C3 = 560
Total number of ways is 350 + 560 = 910.

Eg 11 (N2007/P2/8ai)
Each of seven cards has on it one of the digits 1,2,3,4,5,6,7; no two cards have the same digit. Four of these cards are selected and arranged to form a 4-digit number. (i) How many different 4-digit numbers can be formed in this way?

1st number(choose 1 from 7 numbers) : 7C1 = 7
2nd number(choose 1 from balance 6 numbers): 6C1 = 6
3rd number(choose 1 from balance 5 numbers): 5C1 = 5
4th number(choose 1 from balance 4 numbers): 4C1 = 4
Total number of ways is 7 x 6 x 5 x 4 = 840

Eg 12 (N2007/P2/8b)
4 people are selected to form a debating team from a group of 5 men and 2 women.
(i) Find the number of possible teams that can be selected.
(ii) How many of these teams contain at least 1 woman?

(i) 7C4 = 35
(ii) all men team : 5C4 = 5
35 - 5 = 30

Additional Mathematics - Permutations and Combinations (formulae)

Factorial n
n! = n x (n-1) x (n-2) x (n-3) x ...
Eg, 8!
= 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
= 40,320

Eg, How many ways to arrange 6 books on a shelf:
Ans : 6! ways

Permutations
Arrangement of a list of objects where order does matter.

nPr = n!/[(n-r)!]
Eg 8P3
= 8!/[(8 - 3)!]
= (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / ( 5 x 4 x 3 x 2 x 1)
= 8 x 7 x 6
= 336

Eg How many ways to pick a President, Vice President and Treasurer from a group of 10? The order does matters on who is the President, Vice President or Treasurer.
Ans 10P3 = 720
The permutation formulae denominator removes 4th to 10th order with no title.

Eg List down your 3 favorite desserts, in order, from a menu of 10.
Ans: 10P3 = 720.
The permutation formulae denominator removes the 4th to 10th desert which are not your favourite.

Combinations
Arrangement of a group of objects where order does not matter.

nCr = n!/[(n-r)! x r!]
Eg 8C3
= 8!/[(8-3)! x 3!)
= 8!/(5! x 3!)
= 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 /[(5 x 4 x 3 x 2 x 1) x (3 x 2 x 1]

Eg Picking a team of 3 people from a group of 10. The order is not important.
10C3 = 120

Eg Choose 3 desserts from a menu of 10. The order of which is your favourite is not important.
10C3 = 120.

Although you can use a scientific calculator to derive at the above answers. It is important that you understand the difference in the formulaes of permutation and combination in order to apply it.
= 56

Additional Maths - Permutations & Combinations (Encouragement)

During my school days, A Maths was my favourite subject. But, there was a chapter that I struggled to understand the concept. In fact, the whole class was struggling over this chapter. Later, in my life, I learnt that my friends from other schools have the same problem. So, if you too have problem with this chapter, you can still be good in Additional Maths.

The ShingLee Additional Maths textbook 8th Edition and the Longman Additional Maths guide 3rd edition have omitted this chapter. It was still available in SAP Additional Maths guide 10 years series.

If you are considering giving up this topic, pls think twice. It may miss your chance of scoring distinction. Read my posts on permutations and combainations. I hope it helps.

Contact Miss Koh

If you need help with your school works, pls contact Miss Koh at:

Email : SwallowTuition@gmail.com
HP : 91174497

I provide the following tuition services:
1) Pre-schoolers : Phonics, English and Mathematics
2) Primary levels : Mathematics
3) Secondary levels : Mathematics
4) Secondary levels : Principles of Accounts
5) A levels : Principles of Accounts

The tuition is conducted in the comfort of my condominium residence a in Sengkang Condominium. It is fully equipped with tables and chairs, suitable for use for pre-schoolers and older children, whiteboard, lots of teaching materials and my library of storybooks collection for preschoolers.

If you are a good swimmer, you can dip into the condominium pool to stretch your muscles after the lesson. As there is no lifeguard around, I would require parents' consent before I allow my students to dip into the pool.

I am currently actively on the lookout for suitable space to operate my tuition centre. In the meantime, please make do with my residence. I am able to offer a lower tuition fee as I need not factor in any rental.

I also provide home tuition services.

Welcome to Swallow Tuition Centre.

My Tuition experience

I started tuition when I was in Primary 3. My mother was concerned that I was unable to cope with my school work, and yet, she was unable to help me. Like all mother, she wanted all the best in me. At my end, I dreaded attending tuition. I disliked having an adult watching over my shoulder on everything I did in those evenings, and skipping my favourite television programmes. I do not see the benefit of tuition.
At primary 3, my tutor conducted her lessons at home. There were many children, of different primary levels, cramped into 2 tables in my tutor's kitchen. There were 2 tutors, who were friends, one of whom taught English and another taught Maths. My recollection of those evenings were that I was made to copy lots of problem sums from a primary 4 book. And I were unable to do any of the questions. I do not think I benefited from the lessons.
At primary 4, my tutor was a school teacher. There were about 7 students cramped in my tutor's bedroom. Ironically, my tutor was not at home throughout the year. He returned home at 9pm with a stench of liqour in his breath. We spent our time chatting. Our parents paid our tuition fees blindly.
At primary 5, my mum engaged a home tutor. She taught my 2 sisters, who were in the secondary levels, my brother and myself in the primary levels. I was shy having a visitor at home. I recalled that I sat quietly doing my work all the evenings.
At primary 6, my mum changed the tutor again. He was Edmund. He taught my brother and myself. Both of us were very fond of him. I could still recall him teaching me grammar, explaining to me the questions in my Science and Maths assessment books. After that year, he sent greeting cards to my brother and myself on every Christmas.
I am a tutor now. I want to be a tutor that change my students' results. And hopefully, they have a good memory of me, many years later.

Maths Study Tips

1) Pay attention in class
Do not sleep, dream or talk during lessons. Paying attention in class will save you a lot of time having to figure out the examples yourself.
2) Finish all the exercises your teacher assigns at the end of each lesson
Do your exercises while your teacher explanation is fresh in your mind. Submit your exercises on time gives your teacher a chance to identify any errors and guide you.
3) Get help if you are unable to solve the questions
If you are unable to solve any questions after 5 trials and errors; take a break and return to try it again. If you are still unable to solve the question, consult your teacher, friends and tutor (if any) for help. Do not suffer in silence.
4) Prepare for your tests early
Start preparing for your tests a fortnight in advance. This gives you ample time for practise. When you are preparing for your tests, practise the exercises thorougly so that you can memorise the formulaes and steps. Thereafter, you can try some assessments books.
5) Prepare for your examination early
Start preparing for your examination a month in advance. You can practise the same set of questions again and again until you could memorise them.

All the best to you