Grade 12
Chapter 11
-
Counting Principles
DAY 23 6 May 2020
It is part of probability
Fundamental Counting Principles p309
Rule 1
If one operation can be done in m ways and a second operation can be done in n ways, then the total
possible number of different ways in which both operations can be done is 𝑚 × 𝑛
Example
A coin are tossed and a the is rolled. The possible outcomes are:
K1
S1
K2
S2
K3
S3
K4
S4
K5
S5
K6
S6
2 × 6 = 12 total number of outcomes
Example
A gift pack can be made up as follows:
Choice 1:
Choose one CD out of a possible 4 different CD’s
Choice 2:
Choose one packet of chips out of a possible 5 different types
Choice 3:
Choose one chocolate type out of a possible 12 different types
Choice 4:
Choose one type of fruit out of a possible35 different fruit types
How many different gift packs can be made?
Solution:
4 × 5 × 12 × 3 = 720 different gift packs
Example with repetition
When something can be repeated e.g. letters and digits to create a code
Remember:
There are 26 letters in the alphabet and 0 – 9 consists of 10 digits.
The number of arrangements of n things taken in n ways is 𝑛𝑛
Example
Consider the word PARKTOWN. You are required to form different eight-letter word arrangements using
the letters of the word PARKTOWN. An example APKROTWN. It doesn’t have to make sense. How many
possible arrangements can be made if The letters may be repeated
Solution
An example: AAPKWANO. For each position there are 8 possible letters.
8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 = 88 = 16777216 arrangements
1
Grade 12
Chapter 11
Counting principles
Factorial
The product of 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 can be written as 8! (eight factorial) 0! = 1
Example without repetition
When something can’t be repeated e.g. children sitting in a row and can’t sit on 2 seats at a time
Rule 2
The number of arrangements of n different things taken in n ways is: 𝑛!
Example
On how many ways can 6 vacant places be filled in the first 6 seats of a theatre?
Solution
Benson, Junior, Sage, Caleb, Sipho en Thando moet in ‘n ry sit
Possible
options
Example:
Seat 1
Seat 2
Seat 3
Seat 4
Seat 5
Seat 6
Benson,
Junior,
Sage,
Caleb,
Sipho
Thando
Junior,
Sage,
Caleb,
Sipho
Thando
Sage,
Caleb,
Sipho
Thando
Caleb,
Sipho
Thando
Sipho
Thando
Thando
6 options to
choose
from
5 options to
choose
from
4 options to
choose
from
3 options to
choose
from
2 options to
choose
from
Only 1
option left
Benson
Junior
Sage
Caleb
Sipho
Thando
6
5
4
3
2
1
6 × 5 × 4 × 3 × 2 × 1 = 6! = 720 ways
Grey College
2
Grade 12
Chapter 11
Example
In how many ways can 7 vacant places be filled by 10 different people?
Counting principles
Solution
10 people can occupy 7 places in the following ways:
10 × 9 × 8 × 7 × 6 × 5 × 4
Example
Consider the word PARKTOWN. You are required to form different eight-letter word arrangements using
the letters of the word PARKTOWN. An example APKROTWN. It doesn’t have to make sense. How many
possible arrangements can be made if: The letters may not be repeated
Solution
An example: NOTWARPK. For the first letter there is 8 possibilities, for the 2nd letter there is 7
possibilities etc.
8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 8! = 40 320 arrangements
Example
Consider the word LOVERS
a) How many six letter word arrangements can be made if the letters may be repeated?
b) How many six letter word arrangements can be made if the letters may not repeat?
c) How many 4 letter word arrangements can be made if the letters may be repeated?
d) How many 4 letter word arrangements can be made if the letters may not be repeated?
Solution
a) If letters may repeat, we use exponential notation:
6 × 6 × 6 × 6 × 6 × 6 = 66 = 46656
b) If the letters may not be repeated, we use factorial notation
6 × 5 × 4 × 3 × 2 × 1 = 6! = 720
4
c) 6 = 1296
d) 6 × 5 × 4 × 3 = 360
Grey College
3
Grade 12
Chapter 11
Counting principles
Homework
Exercise 1
p 312
no. b, d, e, g
Grey College
4
Grade 12
Chapter 11
DAY 24
7 May 2020
Grey College
Counting principles
5
Grade 12
Chapter 11
Counting principles
Homework
Exercise 2
p 315
no. a, b
Grey College
6
Grade 12
Chapter 11
Counting principles
DAY 25 8 May 2020
Grouping ~ Arrangements of objects in a row p313
Example
a) In how many ways can 3 boys and 2 girls sit in a row?
b) In how many ways can they sit in a row is a boy and his girlfriend must sit together?
c) In how many ways can they sit in a row if the boys and girls are each to sit together?
d) In how many ways can they sit in a row sit if just the girls are to sit together?
e) In how many ways can they sit in a row sit if just the boys are to sit together?
f) In how many ways can they sit in a row sit if the boys and girls are to alternate?
Solution
a) 5 × 4 × 3 × 2 × 1 = 5! = 120 ways
b)
2! × 4! = 48 ways (2! 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑐𝑜𝑢𝑝𝑙𝑒 𝑎𝑛𝑑 4! 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑤𝑎𝑦𝑠)
c) Boys can sit in 3! ways and the girls in 2! ways.
Together it is 2 × 3! × 2! = 24 ways
d) 2! × 4! = 48 ways (2! 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑔𝑖𝑟𝑙𝑠 𝑎𝑛𝑑 4! 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑤𝑎𝑦𝑠)
e) 3 × 2! × 3! = 36 ways (2! 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑔𝑖𝑟𝑙𝑠 𝑎𝑛𝑑 3! 𝑏𝑜𝑦𝑠)
f) 1 × 2! × 3! = 12 ways (SDSDS)
Grey College
7
Grade 12
Chapter 11
Counting principles
Homework
Exercise 2
p 315
no. c, d, f
Grey College
8
Grade 12
Chapter 11
DAY 26
11 May 2020
Probability
Counting principles
Example
A combination to a lock is formed using three letters of the alphabet, excluding the letters O, Q, S, U, V and
W and using any three digits. The numbers and letters can be repeated. Calculate the probability that a
combination, chosen at random:
a) starts with the letter X and ends with the number 6
b) has exactly one X
c) has one or more number 6 in it.
Solution
a) Let A be the event that a combination starts with the letter X and ends with the number 6.
Since 20 letters and 10 digits can be used, the combination possible will be:
20 × 20 × 20 × 10 × 10 × 10 = 8 000 000
For event A, the number of possibilities is reduced to 1 × 20 × 20 × 10 × 10 × 1 = 40 000
P(A) =
40000
8000000
=
1
200
b) Let B be the event of choosing exactly on X
(1 × 19 × 19 × 10 × 10 × 10) + (19 × 1 × 19 × 10 × 10 × 10) + (19 × 19 × 1 × 10 × 10 × 10)
= 1 083 000
1083000
1083
P(B) = 8000000 = 8000
c) Let C be the event of at least one 6 being chosen
Since 20 letters and 10 digits are used, the number of combinations:
20 × 20 × 20 × 10 × 10 × 10 = 8 000 000
The number of combinations without a six is:
(20 × 20 × 20 × 9 × 9 × 9) = 5832000
The number of combinations with at least one six is:
8000000 - 5832000 = 2168000
2168000
217
P(C) = 8000000 = 1000
Grey College
9
Grade 12
Chapter 11
Counting principles
Homework
Exercise 3
p 316
no. a, d, e, g
Grey College
10
Grade 12
Chapter 11
Counting principles
DAY 27 12 May 2020
Letter arrangement if the letters are identical
Rule 3
The number of different ways which n letters can be arranged where 𝑚1 of the letters are identical, 𝑚2 of
the letters are identical, 𝑚3 of the letters are identical, ...., 𝑚𝑛 of the letters are identical, are:
𝑛!
where the repeated letters are treated as identical
𝑚 !×𝑚 !×𝑚 !×….×𝑚 !
1
2
3
𝑛
Example
Consider the letters in the word NEEDED
a) How many word arrangements can be made with this word if the repeated letters are treated as different
letters?
b) How many word arrangements can be made with this word if the repeated letters are treated as identical
letters?
c) How many word arrangements can be made with this word if the word starts and ends with the same
letter?
Solution
a) 6 × 5 × 4 × 3 × 2 × 1 = 6! = 720 ways
b)
6!
3!×2!
= 60
c)
D
D
4!
=4
3!
NEEE (4 letters and 3 E’s)
E
E
4!
= 12
2!
NDED (4 letters and 2 D’s)
4!
Total number of possible arrangements = 3! +
4!
2!
= 4 + 12 = 16
Grey College
11
Grade 12
Chapter 11
Counting principles
Homework
Exercise 4
p 319
no. b, c, d, e, f
Grey College
12