Gr 12 THIRD PAPER: PROBABILITY
LESSON 4: PERMUTATIONS and COMBINATIONS
Page 1 of 5
YOU ALREADY KNOW THAT:
* The number of arrangements of n different objects in n different ways is n!
(Lesson 3)
Remember: n! reads “n factorial” and n! = n(n-1)(n-2) (n-3)x … x 3 x 2 x 1
PERMUTATIONS:
A permutation is an arrangement of objects in a particular/ spesific order.
In other words permutations are ordered arrangements.
PARTIAL: The number of arrangements/ permutations of r objects chosen
from n unlike (non-identical) objects is:
n!
nP r =
(n r )!
TOTAL: When all the objects (n) are arranged use:
nP n =
n!
(See *)
Remember: 0! = 1
Example 1:
In how many different ways can 10 friends arrive at your party?
(Assume that they arrive 1 at a time!)
Answer:
10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 10! = 3 628 800
or
calculate
10P 10
on your calculator.
(10P 10 = 10!)
(FCP)
Gr 12 THIRD PAPER: PROBABILITY
LESSON 4: PERMUTATIONS and COMBINATIONS
Page 2 of 5
Example 2:
In how many different ways can 6 of your 10 friends arrive at your party?
(Assume that they arrive 1 at a time!)
Answer:
Use FCP: 10 x 9 x 8 x 7 x 6 x 5 = 151 200
or
use nP r - definition: 10P 6 =
10! 3 628 800
10!
=
=
= 151 200
24
(10 6)! 4!
or
use the nP r - key on your calculator:
10P 6
= 151 200
Example 3:
Calculate the number of permutations of a group of 3 objects from a group of
12 different objects.
Answer:
FCP = 12 x 11 x 10 = 1 320
or
definition:
12P 3 =
479 001 600
12!
12!
=
=
= 1 320
362880
9!
(12 3)!
or
calculator: 12P 3 = 1 320
Gr 12 THIRD PAPER: PROBABILITY
LESSON 4: PERMUTATIONS and COMBINATIONS
Page 3 of 5
Example 4:
Calculate the number of distinguishable or distinct permutations that can be
made from all the letters of the word STELLENBERG.
Answer:
[Do not get confused here…… Here you have identical letters. [See lesson 3.]
11!
2! 3!
All the letters nPn.
Answer is divided by repeating letters
i.o.w. the final answer becomes smaller/ less.
2 Ls, 3 Es
= 3 326 400
COMBINATIONS:
A combination is a selection of r objects from n non-identical objects where
the order is not important.
nC r
=
n!
(n r )! r!
For example: The word arrangements SAD and DAS are the same
combination of the letters A, D and S, because order is not important.
(They are different arrangements / permutations! )
Gr 12 THIRD PAPER: PROBABILITY
LESSON 4: PERMUTATIONS and COMBINATIONS
Page 4 of 5
Example 5:
In how many different ways can 5 children be chosen from a group of 30
to pick up litter on the school grounds?
Answer:
30C 5 =
30!
30!
=
= 142 506
(30 5)! 5! 25! 5!
or
use nC r - key on calculator.
(Order not important.
5! )
Example 6:
How many different flags consisting of 3 horizontal pieces or stripes can be
made from material that is available in 8 plain colours, if ….
a) each stripe must have another colour in a specific position?
b) there are no restrictions on the use of these 8 colours?
c) ** each stripe must have another colour and the relative position of the
colour does not matter?
Gr 12 THIRD PAPER: PROBABILITY
LESSON 4: PERMUTATIONS and COMBINATIONS
Page 5 of 5
Answer:
a) FCP: 8 x 7 x 6 = 336
or
use
8P 3
= 336 on calculator.
b) FCP: 8 x 8 x 8 = 83 = 512
c) ** 8C 3 =
8!
8!
=
= 56
(8 3)! 3! 5! 3!
or
use 8C 3 - key on calculator.
Key words:
Arrange
Specific order
Select
No specific order
Permutation
Combination

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