Permutations and Combinations with Probability – HSC Extension 1 Maths Revision

permutations combinations HSC revision extension 1 mathematics

Introduction:

Permutations and combinations are among the most important topics in HSC Extension 1 Mathematics, especially for exam-level probability questions. Many students lose marks due to confusion between permutations and combinations or by applying the wrong formula. In this article, you’ll learn clear concepts, essential formulas, and step-by-step examples to confidently solve permutations, combinations, and probability problems for HSC exams.

Multiplication Rule

If one event can occur in m ways, a second event in n ways, and a third event in r, then the three events can occur in m × n × r ways.

Example: Erin has 5 tops, 6 skirts, and 4 caps from which to choose an outfit. In how many ways can she select one top, one skirt, and one cap?

Solution: Ways = 5 × 6 × 4

Repetition of an Event

If one event with n outcomes occurs r times with repetition allowed, then the number of ordered arrangements is n^r

Example 1 What is the number of arrangements if a die is rolled

(a) 2 times ? 6 × 6 = 6²

(b) 3 times? 6 × 6 × 6 = 6³

(b)r times? 6 × 6 × 6 × ……. = 6^r

Repetition of an Event

Example 2

(a) How many different car number plates are possible with 3 letters followed by 3 digits? Solution: 26 × 26 × 26 × 10 × 10 × 10 = 26³ × 10³

(b) How many of these number plates begin with ABC?

Solution: 1 × 1 × 1 × 10 × 10 × 10 = 10³

(c) If a plate is chosen at random, what is the probability that it begins with ABC?

Solution: 

Factorial Representation

n! = n(n – 1)(n – 2)………..3 × 2 × 1

For example 5! = 5.4.3.2.1 Note 0! = 1

Example a) In how many ways can 6 people be arranged in a row?

Solution: 6.5.4.3.2.1 = 6!

b) How many arrangements are possible if only 3 of them are chosen?

Solution: 6.5.4 = 120

Arrangements or Permutations

Distinctly ordered sets are called arrangements or permutations.

The number of permutations of n objects taken r at a time is given by:

where n = number of objects

          r = number of positions

Arrangements or Permutations

Eg 1. A maths debating team consists of 4 speakers.

a)   In how many ways can all 4 speakers be arranged in a row for a photo?

Solution: 4.3.2.1 = 4! or ⁴P₄               

b)  How many ways can the captain and vice-captain be chosen?

Solution: 4.3 = 12 or ⁴P₂

Arrangements or Permutations

Eg 2. A flutter on the horses There is 7 horses in a race.

a) In how many different orders can the horses finish?

Solution: 7.6.5.4.3.2.1 = 7! or ⁷P₇

b) How many trifectas (1st , 2nd and 3rd ) are possible?

    Solution: 7.6.5 = 210 or ⁷P₃

Permutations with Restrictions

Eg. In how many ways can 5 boys and 4 girls be arranged on a bench if

a) there are no restrictions?

Solution: 9! Or ⁹P₉

c) boys and girls alternate?

Solution: A boy will be on each end

BGBGBGBGB = 5 x 4 x 4 x 3 x 3 x 2 x 2 x 1 x 1

                        = 5! x 4! or ⁵P₅ x ⁴P₄

Permutations with Restrictions

Eg. In how many ways can 5 boys and 4 girls be arranged on a bench if

c) boys and girls in separate groups?

Solution : Boys & Girls or Girls & Boys = 5! x 4! + 4! x 5! = 5! x 4! x 2 or ⁵P_5 x ⁴P_{4 x 2}

d) Anne and Jim wish to stay together?

Solution : (AJ) _ _ _ _ _ _ _ = 2 x 8! or 2 x ⁸P₈

Arrangements with Repetitions

If we have n elements of which x are alike of one kind, y are alike of another kind, z are alike of another kind,

………… then the number of ordered selections or permutations is given by:

Arrangements with Repetitions

Eg.1 How many different arrangements of the word PARRAMATTA are possible?

Solution: 10 letters but note repetition (4 A’s, 2 R’s, 2 T’s)

P

A A A A       No. of arrangements = 

R R                                        = 37 800

M

T T

Arrangements with Restrictions

Eg 1. How many arrangements of the letters of the word REMAND are possible if:

a) there are no restrictions?

Solution: ⁶P₆ = 720 or 6!

b) they begin with RE?

Solution: R E _ _ _ _ = ⁴P₄ = 24 or 4!

c) they do not begin with RE?

Solution: Total – (b) = 6! – 4! = 696

Arrangements with Restrictions

Eg 1. How many arrangements of the letters of the word REMAND are possible if:

d) they have RE together in order?

Solution: (RE) _ _ _ _ = ⁵P₅ = 120 or 5!

e) they have REM together in any order?

Solution: (REM) _ _ _= 3 P3 × 4 P4 = 144

f) R, E, and M are not to be together?

Solution: Total – (e) = 6! – 144 = 576

Arrangements with Restrictions

Eg 2. There are 6 boys who enter a boat with 8 seats, 4 on each side. In how many ways can

a) they sit anywhere?

Solution: ⁸P₆

b) two boys A and B sit on the port side and another boy W sits on the starboard side?

Solution: A & B = ⁴P₂

W = ⁴P₁

Others = ⁵P₃

Total = ⁴P_2 x ⁴P_1 x ⁵P₃

Arrangements with Restrictions

Eg 3. From the digits 2, 3, 4, 5, 6

a) how many numbers greater than 4 000 can be formed?

Solution: 5 digits (any) = ⁵P₅

4 digits (must start with digit >= 4) = ³P₁ x ⁴P₃ Total = ⁵P₅ + ³P₁ x ⁴P₃

b) how many 4-digit numbers would be even?

Even (ends with 2, 4 or 6) = _ _ _ ³P₁ = ⁴P₃ x ³P₁

Circular Arrangements

Circular arrangements are permutations in which objects are arranged in a circle. Consider arranging 5 objects (a, b, c, d, e) around a circular table. The arrangements

abcde                                                                       

bcdea

cdeab

deabc

eabcd

are different in a line, but are identical around a circle

Circular Arrangements

To calculate the number of ways in which n objects can be arranged in a circle, we arbitrarily fix the position of one object, so the remaining (n-1) objects can be arranged as if they were on a straight line in (n-1)! ways.

i.e. the number of arrangements = (n – 1) ! in a circle

Circular Arrangements

Eg 1. At a dinner party, 6 men and 6 women sit at a round table. In how many ways can they sit if:

a) there are no restrictions

    Solution: (12 – 1)! = 11!

b) men and women alternate

    Solution: (6 – 1)! x 6! = 5! x 6!

Circular Arrangements

Eg 1. At a dinner party, 6 men and 6 women sit at a round table. In how many ways can they sit if:

c) Ted and Carol must sit together

   Solution: (TC) & other 10 = 2! x 10!

d) Bob, Ted, and Carol must sit together

   Solution: (BTC) & other 9 = 3! x 9!

Circular Arrangements

Eg 1. At a dinner party, 6 men and 6 women sit at a round table. In how many ways can they sit if:

d) Neither Bob nor Carol can sit next to Ted.

Solution: Seat 2 of the other 9 people next to Ted in (9 x 8) ways or ⁹P₂ Then sit the remaining 9 people (including Bob and Carol) in 9! ways

Ways = (9 x 8) x 9! or ⁹P_{2 x 9!}

Circular Arrangements

Eg 2. In how many ways can 8 differently colored beads be threaded on a string?

Solution: As the necklace can be turned over, clockwise and anti-clockwise arrangements are the same = (8-1)! ÷ 2 = 7! ÷ 2

Unordered Selections

The number of different combinations (i.e. unordered sets) of r objects from n distinct objects is represented by :

                             number of permutations

No.of Combinations = -------------------------------

                             arrangements of r objects

and is denoted by

Combinations

Eg 1. How many ways can a basketball team of 5 players be chosen from 8 players?

Solution: ⁸C₅

Combinations

Eg 2. A committee of 5 people is to be chosen from a group of 6 men and 4 women. How many committees are possible if

a) there are no restrictions?

Solution: ¹⁰C₅

b) one particular person must be chosen for the committee?

Solution: 1 x ⁹C₄

c) one particular woman must be excluded from the committee?

Solution: ⁹C₅

Combinations

Eg 2. A committee of 5 people is to be chosen from a group of 6 men and 4 women. How many committees are possible if:

d) there are to be 3 men and 2 women?

Solution: Men & Women = ⁶C₃ x ⁴C₂

e) there are to be men only?

Solution: ⁶C₅

f) there is to be a majority of women?

Solution: 3 Women & 2 men or 4 women & 1 man = ⁴C₃ x ⁶C₂ + ⁴C₄ x ⁶C₁

Combinations

Eg 3. In a hand of poker, 5 cards are dealt from a regular pack of 52 cards.

(i) What is the total possible number of hands if there are no restrictions?

Solution: ⁵²C₅

Combinations

Eg 3. In a hand of poker, 5 cards are dealt from a regular pack of 52 cards.

ii) In how many of these hands are there:

a) 4 Kings?

Solution: ⁴C₄ x ⁴⁸C₁ or 1 x 48

b) 2 Clubs and 3 Hearts?

Solution: ¹³C₂ x ¹³C₃

_Combinations

Eg 3. In a hand of poker, 5 cards are dealt from a regular pack of 52 cards.

ii) In how many of these hands are there:

c) all Hearts?

Solution: ¹³C₅

d) all the same color?

Solution: Red or Black ²⁶C₅ + ²⁶C₅ = 2 x ²⁶C₅

_Combinations

Eg 3. In a hand of poker, 5 cards are dealt from a regular pack of 52 cards.

ii) In how many of these hands are there:

e) four of the same kind?

Solution: ⁴C₄ x ⁴⁸C₁ x 13 = 1 x 48 x 13

f) 3 Aces and two Kings?

Solution: ⁴C₃ x ⁴C₂

Further Permutations and Combinations

Eg.1 If 4 Maths books are selected from 6 different Maths books and 3 English books are chosen from 5 different English books, how many ways can the seven books be arranged on a shelf:

a) If there are no restrictions?

Solution: ⁶C₄ x ⁵C₃ x 7!

c) If the 4 Maths books remain together?

Solution: = (MMMM)___

            = ⁶P₄ x ⁵C₃ x 4! or (⁶C₄ x 4!) x ⁵C₃ x 4!

Further Permutations and Combinations

Eg.1 If 4 Maths books are selected from 6 different Maths books and 3 English books are chosen from 5 different English books, how many ways can the seven books be arranged on a shelf if:

c) a Maths book is at the beginning of the shelf?

Solution: = M_______

            = 6 x ⁵C₃ x ⁵C₃ x 6!

Further Permutations and Combinations

Eg.1 If 4 Maths books are selected from 6 different Maths books and 3 English books are chosen from 5 different English books, how many ways can the seven books be arranged on a shelf if:

d) Maths and English books alternate

Solution: = M E M E M E M

            = ⁶P₄ x ⁵P₃

Further Permutations and Combinations

Eg.1 If 4 Maths books are selected from 6 different Maths books and 3 English books are chosen from 5 different English books, how many ways can the seven books be arranged on a shelf if:

e) Maths is at the beginning and an English book is in the middle of the shelf.

Solution: M __ E ___

          = 6 x 5 x ⁵C₃ x ⁴C₂ x 5!

Further Permutations and Combinations

Eg 2. (i) How many different 8-letter words are possible using the letters of the word SYLLABUS?

Solution: 2 S's & 2 L's

Words = 

= 10 080

Further Permutations and Combinations

SYLLABUS = 10 080 permutations

(ii) If a word is chosen at random, find the probability that the word:

a) contains the two S’s together

Solution: (SS) ______ (Two L's)

Words = 7!/2! = 2520 Prob = 2520/10080 = 1/4

b) begins and ends with L

Solution: L _______ L (Two S's)

Words = 6!/2! = 360 Prob = 360/10080 = 1/28

Conclusion:

Permutations and combinations are scoring topics in HSC Extension 1 Mathematics when the concepts are clear, and formulas are applied correctly. By understanding whether order matters, choosing the right formula, and practising exam-style questions, you can avoid common mistakes and gain confidence in probability problems. Use this guide as a quick revision before exams and focus on step-by-step working to secure full marks.