grade 12 chapter 11 - counting principles - grey college€¦ · grade 12 chapter 11 counting...

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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 K2 K3 K4 K5 K6 S1 S2 S3 S4 S5 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

Grade 12 Chapter 11 Counting principles

Grey College 2

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

Seat 1 Seat 2 Seat 3 Seat 4 Seat 5 Seat 6 Possible options

Benson, Junior, Sage, Caleb, Sipho Thando 6 options to choose from

Junior, Sage, Caleb, Sipho Thando 5 options to choose from

Sage, Caleb, Sipho Thando 4 options to choose from

Caleb, Sipho Thando 3 options to choose from

Sipho Thando 2 options to choose from

Thando Only 1 option left

Example: Benson Junior Sage Caleb Sipho Thando

6 5 4 3 2 1

6 × 5 × 4 × 3 × 2 × 1 = 6! = 720 ways


Grey College 3

Example In how many ways can 7 vacant places be filled by 10 different people?

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 c) 64 = 1296 d) 6 × 5 × 4 × 3 = 360


Grey College 4

Homework

Exercise 1 p 312 no. b, d, e, g


Grey College 5

DAY 24 7 May 2020


Grey College 6

Homework

Exercise 2 p 315 no. a, b


Grey College 7

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 8

Homework

Exercise 2 p 315 no. c, d, f


Grey College 9

DAY 26 11 May 2020 Probability

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

P(B) = 1083000

8000000=

1083

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

P(C) = 2168000

8000000=

217

1000


Grey College 10

Homework

Exercise 3 p 316 no. a, d, e, g


Grey College 11

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!

𝑚1!×𝑚2!×𝑚3!×….×𝑚𝑛! where the repeated letters are treated as identical

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!

3!= 4

NEEE (4 letters and 3 E’s) E E

4!

2!= 12

NDED (4 letters and 2 D’s)

Total number of possible arrangements = 4!

3!+

4!

2!= 4 + 12 = 16


Grey College 12

Homework

Exercise 4 p 319 no. b, c, d, e, f

grade 12 chapter 11 - counting principles - grey college€¦ · grade 12 chapter 11 counting...

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