What we need to MULTIPLICATION RULE

know about
PERMUTATION?
Permutations for n different objects

ORDER OF
ARRANGEMENT Permutations of n different objects, taking r
objects each time + (Circular Permutations)
IS IMPORTANT
Permutations for n objects involving
identical objects + (Circular permutations)

Permutations with certain conditions

1. MULTIPLICATION RULE
QUESTION 1 :
A shop sells 4 different flavours of yoghurt drink in 3 different sizes.
Nadia wants to buy a bottle of yoghurt drink, find the number of
possible choices.

Flavours, m = 4 Sizes, n = 3

Number of possible choices = m × n

=4×3
= 12
 QUESTION 2 :
There are 3 different routes to travel from location A to location B
and 2 different routes to travel from location B to location C. Find
the number of ways a person can travel from location A to location
C via location B.
Number of routes from A to B, m = 3 Number of routes from B to C, n = 2

Number of ways a person can travel from

location A to location C via location B = m × n
=3×2
= 6
QUESTION 3 :
A Kopitiam shop offer a breakfast set. The menu are as in the diagram
below. Find the number of breakfast set that can be chosen from the
menu if customer need to choose only one from each Menu A, Menu B
and Menu C.

MENU B MENU C
Menu A = 3 Menu B = 2
MENU A
Curry Gravy Coffee
Roti Canai Menu C = 4
Dhal Gravy Milk Tea
Roti Nan Number of breakfast set that can
Milo be chosen = A × B × C
Roti Jala
Nescafe =3×2×4
= 24
2. Permutations for n
different objects
QUESTION 4 :
Without using a calculator, find the value of each of the following.

7! 6!
a) 6! b) c)
3! 3!2!

6! 6 ×5 ×4 ×3 ×2 ×1
a) 6! 7! 7 ×6 ×5 ×4 ×3 ×2 ×1 c)
3!2!
= 3 ×2 ×1 ×2 ×1
b)
3!
= 3×2 ×1
= 6 × 5 × 4 × 3 ×2 × 1
= 720 6 ×5 ×4
=7×6×5×4 =
2 ×1
= 840 = 60
QUESTION 5 :
Arina buys 3 different types of potted plants. Find the number of ways she
can arrange the plants in a row.

3 ways 2 ways 1 ways

Number of ways she can arrange the plants in a row
=3×2×1 3P
=6 3
Ways to arrange the potted plants in a row
QUESTION 6 :
Find the number of 4-digit numbers that can be formed from
the digits 1, 3, 7 and 8 , without repetition of digits

4P
4

Number of 4-digit numbers that can be formed

= 4P4
= 24
 QUESTION 7 :
Find the number of different arrangement of this
nine cards if there is no restrictions

Number of different arrangement of this nine cards

= 9P9
= 362880
 3. Permutations For n Different
Objects ,taking r objects each time
 The number of permutations of
n objects taking r each time is given by
nP 𝑛!
r = , where r ≤ n
𝑛 −𝑟 !

QUESTION 10 :
Without using a calculator, find the value of 5P
3

5P 5!
3 = 5 −3 !

5 × 4 × 3 × 2 ×1
=
2×1

= 5×4×3
= 60
 n=6
QUESTION 11 :
There are 6 parking lots in a row in front of a shop. Find the
number of ways 4 drivers can park their cars in front of the
shop.
r=4
Number of ways 4 drivers can park their cars in front of the shop
= 6P4
6!
= 6−4 !
= 360
QUESTION 12 : n=6 r=4
Given six digits, 1, 2, 3, 4, 5 and 6. Find the number of 4-digit
numbers that can be formed from the 6 digits if repetition of
digits is not allowed
The number of 4 digits numbers that can be formed
= 6P4
6!
=
6 −4 !

= 360
4. Permutations For n Objects
Involving Identical Objects
 The number of permutations for n objects involving identical objects is given by
𝑛!
𝑃=
𝑎!𝑏!𝑐!
where a, b and c, … are the number of identical objects for each type.

QUESTION 15 :
Calculate the number of ways to arrange all the letters of the word
‘ASSESSMENT’.

The number of ways to arrange

10!
= The word ‘ASSESMENT’ consists of 10
4!2!
letters with 4 identical letter ‘S’ and 2
identical letter ‘E’.
= 75600
QUESTION 16 :
There are 5 blue crayons and 3 red crayons in a container. Find the number
of ways to arrange all the crayons in one line.

The number of ways to arrange

8!
= 5!×3!
Identical
= 56
5 blue crayons = 5 !

3 red crayons = 3 !
5. Permutations With Certain
Conditions
QUESTION 17 :
Calculate the number of 3 digit odd numbers that can be formed from
the digits 2,3, 4, 6 and 7 without repetitions.

3 digit odd numbers

4P
The number of
2 3 OR 7 have to end with a
arrangements for the digit 3 or a digit 7
other 4 digits in two
empty spaces is 4P2
The total number of three digit odd
Because we have 2 choices to be numbers
put at 1 place, that is 3 or 7
= 2P1 × 4P2
= 2 × 12
= 24
 QUESTION 18:
3 women and 2 men are lining up to receive free drinks. Find the
number of possible arrangements if two men are at both ends of the
line
We have 2 situation We consider Men 1 = M1
and Men 2 = M2
3P
M1 3 M2 or
The number of possible
M2 3P
3 M1 arrangement
= 3P3 + 3P3
The number of arrangements for the =6+6
other 3 women in three empty = 12
spaces is 3P3
QUESTION 19 :
6 female students and 5 male students are lining up for a photo session.
Find the number of possible arrangements if the female students and
male students have to be arranged alternately.
6 5 5 4 4 3 3 2 2 1 1

6×5×4×3×2×1 5×4×3×2×1
6P
6 × 5P
5
= 86400
QUESTION 20 :
Calculate the number of different ways the letter from the word ‘NUKLEAR’ can
be arranged such that the vowels have to be side by side.

All the 3 vowels can

Vowels = U, E, A Assume that we tied it
together so it will
3P
3
interchange among
become 1 unit themselves

5P Because the vowels become 1 unit,

5 therefore we need to consider to
arrange 5 letter ( N, K, L, R and [U,E,A])

The number of different ways

= 5P5×3P3
= 720
 QUESTION 22 :
9 cards written with the letters from the word “COMMITTEE” are to be arranged
in a row. If the vowels have to be arranged side-by-side, calculate the number of
different arrangements
Arrangement of vowel with identical
Vowels = O, I, E, E Assume that vowels
arranged side by side so it 4!
will become as 1 unit letter ‘E’=
2!
6! 2 identical letter ‘M’ and
Because the vowels become 1 unit,
2! × 2! 2 identical letter ‘T’.
therefore we need to consider to
arrange 6 letter
The number of different arrangements ( C, M, M, T, T and [O,I,E,E] )
6! 4!
= ×
2!×2! 2!
= 180 × 12
= 2160
 QUESTION 23 :
Calculate the number of four digit even number less than 5000 that can be formed
from the digits 3, 4, 5, 6 and 9 without repetitions.

Less than 5000, digits

that can be considered 3 or 4 4 or 6 Even number, digits that can
be considered

Situation A 3 4 or 6
1 × 3P
2
× 2P
1 = 12 No of ways
OR = 12 + 6
Situation B 4 6 = 18
1 3P 1
× 2 × =6
 FORECAST

QUESTION 24:

Given n+2P3 = 30n, find n value. (3 marks)

𝑛!
Formulae nPr = 𝑛 + 2 × 𝑛 + 1 × 𝑛 = 30n
𝑛−𝑟 !
𝑛 + 2 × 𝑛 + 1 = 30
n+2P (𝑛+2)!
3 = = 30n
𝑛+2−3 ! 𝑛2 + 3𝑛 − 28 = 0

(𝑛+2)! 𝑛+7 𝑛−4 = 0

𝑛+2−3 !
= 30n
𝑛 = −7, 𝑛 = 4
𝑛+2 × 𝑛+1 × 𝑛 × 𝑛−1 × 𝑛−2 × 𝑛−3 ×⋯
= 30n ∴𝑛=4
𝑛−1 × 𝑛−2 × 𝑛−3 ×⋯
 FORECAST

QUESTION 25:

(b) Number of arrangement

= 720 – (5P5 × 2P2 )

(4 marks)

= 480
(a)Number of arrangement
Arrangement when P and A
= 6P6 are side by side
= 720 Arrangement with no
condition