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Solve the exercises at the end of chapter 5 – Math 10 Kite
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Solve the exercises Exercise 1 page 20 Math textbook 10 Kite episode 2
a) How many ways are there to arrange 20 students in a vertical line?
A.\({20^{20}}\) B.\(20!\) C. 20 D.1
b) The number of ways to select 3 students from a class of 40 students is:
A. \(A_{40}^3\) B. \({40^3}\) C. \({3^{40}}\) D.\(C_{40}^3\)
Solution method
a) Line up 20 students in a vertical line \( \Rightarrow \) Use the permutation formula
b) Select 3 students from a class of 40 students \( \Rightarrow \) Use combinatorial formula
Solution guide
a) The number of ways to arrange 20 students in a vertical line is: \(20!\) (order). So we choose answer B.
b) The number of ways to select 3 students from a class of 40 students is: \(C_{40}^3\) (choice). So we choose answer D.
Solve the exercises Exercise 2 page 20 Math textbook 10 Kite episode 2
Duong has 2 pairs of pants including: one blue pants and one black pants; 3 shirts include: a brown shirt, a blue shirt and a yellow shirt, 2 pairs of shoes including: a pair of black shoes and a pair of red shoes. Duong wants to choose a set of clothes and a pair of shoes to go sightseeing. By drawing a tree diagram, calculate the number of ways to choose an outfit and a pair of shoes for Duong.
Solution method
Draw a tree diagram by selecting the pants first, then the shirt, and finally the shoes. Then count the number of ways to choose.
Solution guide
Conclusion: From the tree diagram, we see that Duong has 12 ways to choose a set of clothes and a pair of shoes.
Solve the exercise Exercise 3 page 20 Math textbook 10 Kite episode 2
In the plane, give two parallel lines a and b. Given 3 distinct points on line a and 4 distinct points on line b. How many triangles have all 3 vertices at 3 of these 7 points?
Solution method
A triangle is made up of 3 non-collinear points, so to have a triangle we will choose 3 non-collinear points out of 7 given points.
Method 1:
Take 2 points in a, 1 point in b and vice versa
Method 2:
Calculate the number of ways to choose any 3 points from 7 points – the number of ways to choose 3 collinear points on a and b.
Solution guide
Method 1:
TH1: 2 points belong to a and 1 point belongs to b
Number of ways to choose 2 points on line a is \(C_3^2\) (how to choose)
The number of ways to choose a point on line b is: \(C_4^1\) (how to choose)
=> The number of triangles formed is: \(C_3^2 . C_4^1 = 12\)
TH2: 2 points in b and 1 point in a
Number of ways to choose 2 points on line b is \(C_4^2\) (how to choose)
The number of ways to choose a point on the line a is: \(C_3^1\) (how to choose)
=> The number of triangles formed is: \(C_4^2 + C_3^1 = 18\)
So there are 12 + 18 = 30 triangles in all.
Method 2:
Number of ways to choose 3 points on line a is: \(C_3^3\) (how to choose)
Number of ways to choose 3 points on line b is: \(C_4^3\) (how to choose)
Number of ways to choose any 3 points out of 7 given points is: \(C_7^3\) (how to choose)
The number of ways to choose 3 noncollinear points out of 7 given points is: \(C_7^3 – C_4^3 – C_3^3 = 30\) (how to choose)
So the number of possible triangles is : 30 (triangles)
Solve the exercises Lesson 4, page 20, Math textbook 10 Kite episode 2
In the plane, let 6 parallel lines and 8 parallel lines are perpendicular to those 6 lines. How many rectangles are formed?
Solution method
Step 1: Calculate the number of ways to choose 2 parallel lines out of 6 parallel lines
Step 2: Calculate the number of ways to choose 2 parallel lines out of 8 parallel lines that are perpendicular to the original 6 parallel lines
Step 3: Apply the multiplication rule
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Solution guide
Number of ways to choose 2 parallel lines out of 6 parallel lines is: \(C_6^2\) (how to choose)
Number of ways to choose 2 parallel lines out of 8 parallel lines that are perpendicular to the original 6 parallel lines is: \(C_8^2\) (how to choose)
Applying the multiplication rule, we have the number of rectangles that can be formed: \(C_8^2.C_6^2 = 420\) (rectangle)
Solve the exercise Exercise 5 page 20 Math textbook 10 Kite episode 2
Expand the following expressions:
a) \({\left( {4y – 1} \right)^4}\)
b) \({\left( {3x + 4y} \right)^5}\)
Solution method
a) Use Newton’s binomial expansion with \(n = 4\): \({\left( {a + b} \right)^4} = {a^4} + 4{a^3}b + 6{a^2}{b^2} + 4a{b^3} + {b^4}\)
b) Use Newton’s binomial expansion with \(n = 5\):\({\left( {a + b} \right)^5} = {a^5} + 5{a^4}b + 10{a^3}{b^2} + 10{a^2}{b^3} + 5a{b^4} + {b^5}\)
Solution guide
a) \({\left( {4y – 1} \right)^4} = {\left[ {4y + \left( { – 1} \right)} \right]^4} = 256{y^4} – 256{y^3} + 96{y^2} – 16y + 1\)
b) \({\left( {3x + 4y} \right)^5} = 243{x^5} + 1620{x^4}y + 4320{x^3}{y^2} + 5760{x ^2}{y^3} + 3840x{y^4} + 1024{y^5}\)
Solving exercises Lesson 6 page 20 Math textbook 10 Kite episode 2
A computer password is a sequence of characters (in order from left to right) chosen from: 10 digits, 26 lowercase letters, 26 uppercase letters, and 10 special characters. Ngan wants to create a computer password with a length of 8 characters including: the first 4 characters are 4 different digits, the next 2 characters are lowercase letters, the next 1 character. Another is a capital letter, the last character is a special character. How many ways do you have to set up a computer password?
Solution method
Step 1: Select the first 4 characters as 4 different digits from 10 digits (with sorted)
Step 2: Select next 2 characters from 26 lowercase letters
Step 3: Choose the next 1 character from 26 uppercase letters
Step 4: Choose the last 1 character from 10 special characters
Step 5: Apply the multiplication rule
Solution guide
+) The number of ways to choose the first 4 characters is: \(A_{10}^4\) (how to choose)
+) The number of ways to choose the next 2 characters is: \(C_{26}^1.C_{26}^1\) (how to choose)
+) The number of ways to choose the next 1 character is: \(C_{26}^1\) (how to choose)
+) The number of ways to choose the last 1 character is: \(C_{10}^1\) (how to choose)
+) Applying the multiplication rule, we have the number of possible passwords:
\(A_{10}^4.C_{26}^1.C_{26}^1.C_{26}^1.C_{10}^1\) (password)
Solving exercises Exercise 7 page 20 Math textbook 10 Kite episode 2
A high school organized a relay race between classes with the content of 4 x 100 m and required each team to consist of 2 boys and 2 girls. An was assigned by the teacher to select 4 students and arrange their running order to register for the contest. How many ways do you An create a qualifying competition team? Know that An’s class has 22 boys and 17 girls.
Solution method
Step 1: Choose any 2 boys from 22 boys
Step 2: Choose any 2 girls from 17 girls
Step 3: Arrange the 4 you have chosen in some order
Step 4: Apply the multiplication rule
Solution guide
+) Number of ways to choose any 2 boys from 22 boys is: \(C_{22}^2\) (how to choose)
+) Number of ways to choose any 2 girls from 17 girls is: \(C_{17}^2\) (how to choose)
+) The number of ways to arrange the competition order of 4 friends is: \(4!\) (sorting)
+) Applying the multiplication rule, we have the number of ways to form a team: \(C_{22}^2.C_{17}^2.4!\) (how to make )
Solve exercises Exercise 8 page 20 Math textbook 10 Kite episode 2
Uncle Thao wants to buy 2 computers for work. The salesman introduced you to 3 computer manufacturers for your reference: the first company has 4 suitable computers, the second company has 5 suitable computers, the third company has 7 types of suitable computers. How many ways do you have to choose 2 computers for work?
Solution method
Choose any 2 computers from 4+5+7=16 computers => convolution 2 of 16.
Solution guide
+) Total number of matching computers is : \(4 + 5 + 7 = 16\) (computers)
+) The number of ways to choose 2 computers from 16 matching computers is: \(C_{16}^2 = 120\) (how to choose)
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