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**Solving exercises at the end of chapter 8 (Math 10 – Connection textbook) – KNTT**

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A – Quiz

**Lesson 8.17:** The number of ways to put 4 different flowers in 4 different vases (each flower in one vase) is

A. 16

B. 24

C. 8

D. 4

Solution method – See details

Apply the permutation formula.

Detailed explanation

Each arrangement of 4 flowers in 4 different vases (each flower in a vase) is a permutation of 4 elements.

Number of flower arrangements is: 4!= 24

Select REMOVE**Lesson 8.18**:Number of different three-digit numbers where all digits are greater than 0 and less than or equal to 5 is

A. 120

B. 60

C. 720

D. 2

Solution method – See details

Apply the matching formula.

Detailed explanation

From the digits 1, 2, 3, 4, 5, a natural number with 3 different digits is::\(A_5^3 = 60\)

Choose B.**Lesson 8.19:** The number of ways to choose 3 students to take swimming lessons from a group of 10 students is

A. 3 628 800

B. 604 800

C. 120

D. 720

Solution method – See details

Apply the combinatorial formula.

Detailed explanation

Number of ways to choose 3 out of 10 students is: \(C_{10}^3 = 120\)

Choose C.**Lesson 8.20**: Friend An hands over a dice twice. The number of cases where the total number of dots appearing on the dice is 8 over two rolls is

A. 36

B. 6

C. 5

D. 4

Detailed explanation

The cases where the total number of dots appearing on the dice is equal to 8 through two rolls are: (4,4), (3,5), (5,3), (2,6), (6,2) .

Choose C.**Lesson 8.21**: The coefficient of \({x^4}\) in the binomial expansion \({(3x – 4)^5}\) is

A. 1620

B. 60

C. -60

D. -1620

Solution method – See details

Apply the expansion formula of \({(a + b)^5} = {a^5} + 5{a^4}b + 10{a^3}{b^2} + 10{a^2 }{b^3} + 5a{b^4} + {b^5}\)

Detailed explanation

We have:

\(\begin{array}{l}{(3x – 4)^5} = {(3x)^5} + 5{(3x)^4}( – 4) + 10{(3x)^3}{ ( – 4)^2}\\ + 10{(3x)^2}{( – 4)^3} + 5.3x{( – 4)^4} + {( – 4)^5}\end{array }\)

The coefficient of \({x^4}\) in the binomial expansion \({(3x – 4)^5}\) is \({5.3^4}( – 4) – 1620\)

Choose D.

## B. Essay:

### Solve problem 8.22 page 76 Math textbook 10 Connecting knowledge volume 2

a) How many ways are there to write a sequence of 5 uppercase letters from the English alphabet (consisting of 26 letters)?

b) How many different ways of writing a sequence of 5 uppercase letters from the English alphabet (consisting of 26 letters)?

**Solution method**

a) Use the multiplication rule

b) Sorting 5 letters from 26 letters is a 5 of 26 elements convolution

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**Detailed explanation**

a) Since the letters do not need to be different, the number of ways to choose is: 26.26.26.26.26 = 26^{5} = 11 881 376 ways.

b) Selecting and arranging 5 letters from 26 letters is a 5 convolution of 26 elements, so the number of ways is: \(A_{26}^{5}\) = 7 893 600 ways.

### Solve problem 8.23 page 76 Math textbook 10 Connecting knowledge volume 2

From the digits: 1; 2; 3; 4; 5; 6.

a) How many different three-digit numbers can be made?

b) How many different three-digit numbers can be formed that are divisible by 3?

**Solution method**

a) Making a different 3-digit number is taking 3 elements from a given set of digits and then sorting

b) We have triplets of numbers that are divisible by 3

Using the permutation formula

**Detailed explanation**

a) The number of ways is: \(A_{6}^{3}\) = 120 ways.

b) If a number is divisible by 3, then the sum of digits must be divisible by 3. We have triples: (1; 2; 3), (1; 2; 6), (1; 3; 5), (1 ; 5; 6), (2; 3; 4), (2; 4; 6), (3; 4; 5), (4; 5; 6)

Each trio has 3! arrangement

So the number of different ways to create a 3-digit number that is divisible by 3 is: 8.3! = 48 ways.

### Solve problem 8.24 page 76 Math textbook 10 Connecting knowledge volume 2

Cell A has 2n = 8 chromosomes, and mitosis 5 times in a row. B cells have 2n = 14 chromosomes and divide 4 times in a row. Calculate and compare the total number of chromosomes in cell A and in cell B produced.

**Solution method**

– Calculate the number of cells after 5 times of mitosis => number of chromosomes in the cell

– Calculate the number of cells after 4 mitosis => number of chromosomes in the cell

– Compare the number of cells after two mitosis

**Detailed explanation**

+) After 5 times of mitosis, the number of A cells is: 2^{5} = 32 cells.

Infer: the number of chromosomes in cell A produced is: 32.8 = 256 chromosomes.

+) After 4 times of mitosis, the number of B cells is: 2^{4} = 16 cells.

Infer: the number of chromosomes in the B cell produced is: 16.14 = 224 chromosomes.

The total number of chromosomes in cell A is greater than in cell B.

### Solve problem 8.25 page 76 Math textbook 10 Connecting knowledge volume 2

Class 10B has 40 students including 25 boys and 15 girls. How many ways are there to choose 3 students to join the school’s volunteer team in each of the following cases?

a) Any three students are selected.

b) Three students are selected including 1 male and 2 female.

c) There is at least one male out of the three selected students.

**Solution method**

a) Choose any 3 out of 40 students as a convolution of 40 elements

b) Calculate the number of ways to choose 1 male from 25 males

Calculate the number of ways to choose 2 girls from 15 girls

c) Calculate the number of ways to choose. If no male students are selected, then 3 girls will be selected

Calculate the number of ways to choose so that out of the 3 chosen friends, there is at least 1 male friend

**Detailed explanation**

a) Choose any 3 students out of 40 students is a convolution of 40 elements, so the number of ways to choose is: \(C_{40}^{3}\) = 9880 ways.

b) Choose 1 male from 25 males, number of ways to choose: \(C_{25}^{1}\) = 25 ways.

Choose 2 girls from 15 girls, number of ways to choose: \(C_{15}^{2}\) = 105 ways.

So the number of ways to choose 1 male and 2 female is: 25,105 = 2625 ways.

c) Consider the case, no male students are selected, then 3 girls will be selected, the number of ways to choose is: $C_{15}^{3}$ = 455 ways.

In order to have at least 1 male friend out of 3, the number of ways to choose is: $C_{40}^{3}-C_{15}^{3}$ = 9425 ways.

### Solve problem 8.26 page 76 Math textbook 10 Connecting knowledge volume 2

In the Newton binomial expansion of \({\left( {2x + 3} \right)^5}\), the coefficient of x^{4} or coefficient of x^{3} bigger?

**Solution method**

Find the term containing x^{4 }x^{3}

**Detailed explanation**

Term containing x^{4 }in expansion is: 5.(2x)^{4}.3 = 240x^{4}

Derive the coefficient of x^{4} is: 240

Term containing x^{3} in expansion is: 10.(2x)^{3}3^{2 }=720x^{3}

Derive the coefficient of x^{3} is 720.

So the coefficient of x^{4 }greater than the coefficient of x^{3}.