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## Solve Exercise 3. Combination (C5 – Math 10 Kite)

### Solution of Exercises Lesson 1, page 17, Math Textbook 10, Kite episode 2

Given 8 points such that no 3 points are collinear. How many triangles with 3 vertices are 3 out of 8 given points?

**Solution method**

+) A triangle is made up of 3 non-collinear points, so to have a triangle we will choose 3 out of 8 non-collinear points.

=> triangle number is a convolution of 3 of 8 elements.

**Solution guide**

The number of triangles with 3 vertices is 3 of the given 8 points which is a convolution of 3 of 8 elements, so the number of triangles is: \(C_8^3\) (triangle)

### Solve the exercise Exercise 2 page 17 Math textbook 10 Kite episode 2

There are 10 teams participating in a Soccer Exercise Tournament. How many ways are there to arrange a round-robin match so that the two teams meet exactly once?

**Solution method**

+) To have a match, there must be 2 teams participating. Therefore, to have a match, we will choose 2 teams out of 10 teams.

+) So the number of matches is a convolution 2 of 10 elements.

**Solution guide**

The number of ways to arrange the score round match so that the two teams only meet exactly once is a convolution 2 of 10 elements, so the number of ways to arrange the match is: \(C_{10}^2 = 45\) ( arrangement)

### Solution of Exercises Lesson 3, page 17, Math Textbook 10, Kite episode 2

Grade 10 has 16 girls and 18 boys participating in the Green Summer volunteering. The school delegation plans to set up a tree planting team consisting of 3 students, both male and female. How many ways are there to set up such a tree nest?

**Solution method**

*) Analysis: (Number of ways to choose 3hs including both men and women in 34hs) + (Number of ways to choose 3hs for men in 34hs) + (Number of ways to choose 3hs for women in 34hs) = Number of ways to choose any 3hs in 34hs.

+) Step 1: Calculate the number of ways to choose any 3hs in 34hs.

+) Step 2: Calculate the number of ways to choose 3 male students in 34 hours.

+) Step 3: Calculate the number of ways to choose 3 female students in 34 hours.

+) Step 4: Consider the difference to calculate the number of ways to choose 3hs including both men and women in 34hs.

**Solution guide**

+) The number of ways to choose any 3hs in 34hs is: \(C_{34}^3\) (how to choose)

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+) Number of ways to choose 3 male students in 34 hours is: \(C_{18}^3\) (choose method)

+) The number of ways to choose 3 female students in 34 hours is: \(C_{16}^3\) (how to choose)

+) The number of ways to choose 3hs including both men and women in 34hs is: \(C_{34}^3 – C_{18}^3 – C_{16}^3 = 4608\) (how to choose)

### Solving Exercises Lesson 4, page 17, Math Textbook 10 Kite, episode 2

A small shop selling flowers has 50 roses and 60 daisies. Uncle Ngoc wants to buy 5 flowers including both of the above flowers. How many ways does Uncle Ngoc have to choose flowers?

**Solution method**

*) Analysis: ( Number of ways to choose 5 flowers including 2 types out of 110 flowers) + (Number of ways to choose 5 roses out of 50 roses) + (Number of ways to choose 5 daisies out of 60 daisies) = Number How to choose any 5 flowers out of 110 flowers.

+) Step 1: Calculate the number of ways to choose any 5 flowers out of 110 flowers.

+) Step 2: Calculate the number of ways to choose 5 chrysanthemums out of 60 chrysanthemums.

+) Step 3: Calculate the number of ways to choose 5 roses out of 50 roses.

+) Step 4: Consider the difference to calculate the number of ways to choose 5 flowers including 2 types out of 110 flowers.

**Solution guide**

+) The number of ways to choose any 5 flowers out of 110 flowers is: \(C_{110}^5\) (how to choose)

+) Number of ways to choose 5 chrysanthemums out of 60 chrysanthemums is: \(C_{60}^5\) (how to choose)

+) The number of ways to choose 5 roses out of 50 roses is: \(C_{50}^5\) (how to choose)

+) The number of ways to choose 5 flowers including 2 types out of 110 flowers is: \(C_{110}^5 – C_{60}^5 – C_{50}^5\) (how to choose)

### Solve the exercise Exercise 5 page 17 Math textbook 10 Kite episode 2

Sum \(C_{15}^{12} + C_{15}^{13} + C_{16}^{14}\)

**Solution method**

\(C_{n – 1}^{k – 1} + C_{n – 1}^k = C_n^k\left( {1 \le k < n} \right)\)

**Solution guide**

We have: \((C_{15}^{12} + C_{15}^{13} )+ C_{16}^{14} = C_{16}^{13} + C_{16}^{14 } = C_{17}^{14} = 680\)