Permutations & Combinations HSC Revision Extension 1 Mathematics

permutations combinations HSC revision extension 1 mathematics

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

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 x⁴P_{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)

A A A A No. of arrangements =

factorial_problems_chaitu_informative_blogs

R R = 37 800

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 x⁴P_1 x⁵P₃

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 ⁹P₂Then 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 = ⁴C₃x ⁶C₂+ ⁴C₄x 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 =

permutation_combinations_chaitu_informative_blogs

= 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

Permutations & Combinations HSC Revision Extension 1 Mathematics