adsense

**Solve Exercises Lesson 23: Counting rules (Math 10 – Textbook Connecting knowledge)**

—————

### Solve problem 8.1, page 65, Math textbook 10, Connecting knowledge volume 2

On the bookshelf there are 8 books of short stories, 7 novels and 5 volumes of poetry (all different). Draw a tree diagram and show how many ways Phong can choose a book to read on the weekend.

Solve problem 8.1, page 65, Math textbook 10, Connecting knowledge volume 2

On the bookshelf there are 8 books of short stories, 7 novels and 5 volumes of poetry (all different). Draw a tree diagram and show how many ways Phong can choose a book to read on the weekend.

**Solution method**

Use the addition rule

**Detailed explanation**

The number of ways to choose a book to read is: 8 + 7 + 5 = 20 books.

### Solve problem 8.2, page 65 Math textbook 10 Connecting knowledge volume 2

A person tosses a double-sided coin, and after each toss, records whether the result is heads or tails. If the person sows 3 times, how many possibilities are there?

Solve problem 8.2, page 65 Math textbook 10 Connecting knowledge volume 2

A person tosses a double-sided coin, and after each toss, records whether the result is heads or tails. If the person sows 3 times, how many possibilities are there?

**Solution method**

+ Possibility of happening when 1st sowing

+ Possibility of 2nd seeding

+ Possibility of 3rd seeding

Use the multiplication rule

**Detailed explanation**

The first sowing can appear heads or tails, so the number of possibilities is: 2.

Sowing the second time is similar to the first time, the number of possibilities is: 2.

The 3rd seed is similar to the above, the number of possibilities is: 2.

So after 3 times sowing, the number of possibilities is: 2.2.2 = 8.

### Solve lesson 8.3, page 65, Math 10 Textbook, Connecting knowledge, volume 2

In a plant species, A is the dominant gene for double flowers and a is the recessive gene for single flowers.

a) How many genotypes are produced by the combination of these two genes? Write the genotypes.

b) In random mating, how many different mating patterns are there from those genotypes?

Solve lesson 8.3, page 65, Math 10 Textbook, Connecting knowledge, volume 2

In a plant species, A is the dominant gene for double flowers and a is the recessive gene for single flowers.

a) How many genotypes are produced by the combination of these two genes? Write the genotypes.

b) In random mating, how many different mating patterns are there from those genotypes?

**Solution method**

The addition rule is applied when the work is divided into distinct alternatives (do one of the alternatives to complete the job).

The multiplication rule is applied when the job has many successive stages (all stages must be performed to complete the job).

**Detailed explanation**

a) The combination creates 3 genotypes: AA, Aa, aa.

b) When randomly assigning coupons, AA can be generated with AA, Aa, aa.

It follows that there are types: AA ×AA; AA×Aa; AA×aa; Aa×Aa; Aa×aa; aa×aa

There are 6 different mating patterns from those genotypes.

### Solve problem 8.4 page 65 Math textbook 10 Connecting knowledge volume 2

How many natural numbers are there?

a) have 3 different digits?

b) is an odd number with 3 different digits?

c) is a 3-digit number that is divisible by 5?

d) is a 3 digit number that is different and divisible by 5 ?

Solve problem 8.4 page 65 Math textbook 10 Connecting knowledge volume 2

How many natural numbers are there?

a) have 3 different digits?

b) is an odd number with 3 different digits?

c) is a 3-digit number that is divisible by 5?

d) is a 3 digit number that is different and divisible by 5 ?

**Solution method**

The natural number to be formed has the form \(\overline{abc}\), with \(a,b,c \in \left\{ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} \right\}(a \ne 0,{\rm{ }}a \ne b \ne c)\)

+) Find the number of ways to choose the digit a.

adsense

+) Find the number of ways to choose the digit b.

+) Find the number of ways to choose the digit c.

+) Use the multiplication rule.

**Detailed explanation**

a) Call the natural number to be formed of the form: \(\overline{abc}\), where a, b, c belong to the set of numbers A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, (a \(\neq \) 0, \(a\neq b\neq c\)).

There are 9 ways to choose the number a, since a \(\neq \) 0.

Choose b with 9 ways from set A\{a}

Choose c with 8 ways from set A\{a; b}

The number of numbers that satisfy the problem is: 9.9.8 = 648 numbers.

b) Call the natural number to be formed in the form: \(\overline{abc}\), where a, b, c belong to the set of numbers A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, (a \(\neq \) 0, \(a\neq b\neq c\)).

For \(\overline{abc}\) to be odd, then c belongs to the set {1; 3; 5; 7; 9},

Choose c with 5 ways from set {1; 3; 5; 7; 9},

Choose a with 8 ways from the set A\{c; 0}

Choose b with 8 ways from the set A\{c; a}

The number of numbers that satisfy the problem is: 5.8.8 = 320 numbers.

c) Call the natural number to be formed of the form: \(\overline{abc}\), where a, b, c belong to the set of numbers A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, (a \(\neq \) 0)

For \(\overline{abc}\) to be divisible by 5, then c belongs to the set {0; 5},

Choose c in 2 ways from the set {0; 5},

Choose a with 9 ways from set A\{0}

Choose b with 10 ways from set A

So the number of 3-digit numbers that are divisible by 5 is: 2.9.10 = 180 numbers.

d) Call the natural number to be formed of the form: \(\overline{abc}\), where a, b, c belong to the set of numbers A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, (a \(\neq \) 0, \(a\neq b\neq c\)).

For \(\overline{abc}\) to be divisible by 5, then c belongs to the set {0; 5},

+ If c = 0, then: choose a has 9 ways, choose b has 8 ways

=> The number of different 3-digit numbers that end in 0 is: 9.8 = 72 numbers.

+ If c = 5 then: choose a has 8 ways, choose b has 8 ways

=> The number of different 3-digit numbers ending in 5 is: 8.8 = 64 numbers.

So the number of different 3-digit numbers that are divisible by 5 is: 72 + 64 = 136 numbers.

### Solve lesson 8.5, page 65, Math textbook 10, Connecting knowledge volume 2

a) The password of a computer program is specified with 3 characters, each letter is a digit. How many different passwords can be generated?

b) If the computer program specifies a new password, the password still consists of 3 characters, but the first character must be a capital letter in the English alphabet consisting of 26 letters (from A to Z) and the following 2 characters are digits (from 0 to 9). How many different passwords can the new rule create than the old one?

Solve lesson 8.5, page 65, Math textbook 10, Connecting knowledge volume 2

a) The password of a computer program is specified with 3 characters, each letter is a digit. How many different passwords can be generated?

b) If the computer program specifies a new password, the password still consists of 3 characters, but the first character must be a capital letter in the English alphabet consisting of 26 letters (from A to Z) and the following 2 characters are digits (from 0 to 9). How many different passwords can the new rule create than the old one?

**Solution method**

The natural number to be formed is of the form \(\overline{abc}\), with \(a,b,c \in \left\{ {0,1,2,3,4,5,6,7,8, 9} \right\}\), (\(a \ne 0,a \ne b \ne c\))

+) Find the number of ways to choose the digit a.

+) Find the number of ways to choose the digit b.

+) Find the number of ways to choose the digit c.

+) Use the multiplication rule.

**Detailed explanation**

a) Call the natural number to be formed of the form: \(\overline{abc}\), where a, b, c belong to the set of numbers A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Choose a with 10 ways from set A,

Choose b with 10 ways from set A,

Choose c with 10 ways from set A,

So the number of passwords can be generated: 10.10.10 = 1000 passwords.

b) Choose the first letter from the set of 26 letters from A to Z, there are 26 ways to choose,

Choose the second character as a digit with 10 choices,

Choose the third character as a digit with 10 choices.

The number of ways to create a new password is: 26.10.10 = 2600 passwords.

So it is possible to create more than the old rule, the number of passwords is: 2600 – 1000 = 1600 passwords.