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.

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