Permutation
Combination
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 nr
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 × ……. = 6r
Example 2
(a) How many different car number plates are
possible with 3 letters followed by 3 digits?
3 3
Solution: 26 × 26 × 26 × 10 × 10 × 10 = 26 x 10
(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: 103 = 1
263 × 103 263
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
Permutations
Distinctly ordered sets are called arrangements or
permutations.
The number of permutations of n objects taken r at
a time is given by:
n
Pr = n!
( n – r)!
where n = number of objects
r = number of positions
Ex 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?
4
Solution : 4.3 = 12 or P2
Ex 2. A flutter on the horses
There are 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?
7
Solution : 7.6.5 = 210 or P3
Ex 3. In how many ways can 5 boys and 4 girls
be arranged on a bench if
a) there are no restrictions?
9
P9
Solution : 9! or
c) boys and girls alternate?
Solution : A boy will be on each end
BGBGBGBGB = 5 4 4 3 3 2 2 1 1
= 5! 4! or P5 4P4
5
Ex 3. In how many ways can 5 boys and 4 girls
be arranged on a bench if
c) boys and girls are in separate groups?
Solution : Boys Girls or Girls Boys
= 5! 4! + 4! 5! = 5! 4! 2
or 5P5 4P4 2
d) Anne and Jim wish to stay together?
Solution : (AJ) _ _ _ _ _ _ _
= 2 8! or 2 8P8
Arrangements
x
If we have n elements of which 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:
n!
x! y! z!
Ex 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
AAAA
No. of 10!
RR arrangements = 4! 2! 2!
M = 37 800
TT
Ex 2. How many arrangements of the letters of
the word REMAND are possible if:
a) there are no restrictions?
Solution : 6
P6 = 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
Ex 3. 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) _ _ _ = 3P3 × 4P4 = 144
f) R, E and M are not to be together?
Solution : Total – (e) = 6! – 144 = 576
Ex 4. There are 6 boys who enter a boat with 8
seats, 4 on each side. In how many ways can
a) they sit anywhere?
8P6
Solution :
b) two boys A and B sit on the port side and another
boy W sit on the starboard side?
Solution : A B = 4P2
W = 4P1
Others = 5P3
Total = 4P2 4P1 5P3
Ex 5. From the digits 2, 3, 4, 5, 6
a) how many numbers greater than 4 000 can be
formed?
5
Solution : 5 digits (any) = P5
4 digits (must start with digit 4) = 3P1 4P3
Total = 5P5 + 3P1 4P3
b) how many 4 digit numbers would be even?
Even (ends with 2, 4 or 6) = _ _ _ 3P1
= 4P3 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
a abcde
bcdea
e b cdeab
deabc
d eabcd
c
are different in a line, but are identical around a
circle.
a
To calculate the number of ways
in which n objects can be
e b arranged in a circle, we arbitrarily
fix the position of one object, so
the remaining (n-1) objects can
d be arranged as if they were on a
c straight line in (n-1)! ways.
i.e. the number of arrangements = (n – 1) !
in a circle
Ex 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)! 6! = 5! 6!
Ex 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! 10!
d) Bob, Ted and Carol must sit together
Solution : (BTC) other 9 = 3! 9!
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 8) ways or 9P2
Then sit the remaining 9 people
(including Bob and Carol) in 9!
ways
Ways = (9 8) 9! or P2 9!
9
Ex 2. In how many ways can 8 differently
coloured beads be threaded on a string?
Solution :
As necklace can be turned over, clockwise and
anti-clockwise arrangements are the same
= (8-1)! 2 = 7! 2
Combination
The number of different combinations (i.e. unordered
sets) of r objects from n distinct objects is represented
by :
No. of = number of permutations
Combinations
arrangements of r objects
and is denoted by
n n
Cr = Pr = n! _
r! r! ( n – r)!
Ex 1. How many ways can a basketball team of 5
players be chosen from 8 players?
Solution :
8
C5
Ex 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 : 10
C5
b) one particular person must be chosen on the
committee?
Solution : 1 9C 4
c) one particular woman must be excluded from the
committee?
Solution : 9
C5
Ex 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 4C2
e) there are to be men only?
Solution : 6
C5
f) there is to be a majority of women?
Solution :
3 Women 2 men Or 4 Women 1 man
= 4C3 6C2 + 4C4 6C1
Ex 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 :
52
C5
Ex 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 : C4 48C1 or 1 48
4
b) 2 Clubs and 3 Hearts?
Solution :
13
C2 13C3
Ex 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 : 13
C5
d) all the same colour?
Solution : Red or Black 26
C5 + 26C5 = 2 26
C5
=
Ex 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 :
4
C4 48C1 13 = 1 48 13
f) 3 Aces and two Kings?
Solution : C3 4C2
4
Permutations & Combinations
Ex.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 5C3 7!
c) If the 4 Maths books remain together?
Solution : = (MMMM) _ _ _
= 6P4 5C3 4! or ( 6C4 4!) 5C3 4!
Ex.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 5C3 5C3 6!
Ex.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 : = MEMEMEM
= 6P4 5P3
Ex.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) A Maths is at the beginning and an English book
is in the middle of the shelf.
Solution : M__ E___
= 6 5 5C3 4C2 5!
Ex 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 = 8!__
2! 2!
= 10 080
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!_ = 2520 Prob = 2520 = 1
2! 10080 4
b) begins and ends with L
Solution : L______ L (Two S’s)
Words = 6!_ = 360 Prob = 360 = 1
2! 10080 28
