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**Solve Exercises Lesson 24: Permutation, alignment and combination (Math 10 – Connection textbook) – Information skills**

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### Solve problem 8.6 page 70 Math textbook 10 Connecting knowledge volume 2

An artist needs to display 10 different artworks in a row. How many ways can the artist arrange the pictures?

**Solution method**

Use the permutation formula: \(P_n = n(n – 1)(n – 2)…2 . 1 = n!\)

**Detailed explanation**

Sorting 10 pictures in a row is a permutation of 10 elements, so the number of ways to arrange is: 10! = 3 628 800 ways.

### Solve problem 8.7 page 70 Math textbook 10 Connecting knowledge volume 2

From the digits 0, 1, 2, 3, 4, how many different three-digit natural numbers can be formed?

**Solution method**

Use the matching formula: \(A_n^k = n(n – 1)…(n – k + 1) =\frac{n!}{(n – k)!} \) \((1 ≤ k) n)\)

**Detailed explanation**

Creating 3 natural numbers from the set of digits 0, 1, 2, 3, 4 is convolution 3 of 5 words, so the number of ways is \(A_{5}^{3}= 60\) .

However, for 3-digit numbers, hundreds must be non-zero, for numbers of the form \(\overline{0ab}\), the number of ways to form is: \(A_{4}^{2}= 12\) .

So the number of natural numbers with three different digits, made from the digits 0, 1, 2, 3, 4 is: 60 – 12 = 48 numbers.

### Solve problem 8.8, page 70, Math 10 Textbook, Connecting knowledge volume 2

How many ways are there to choose a set of two positive integers less than 100? How many ways are there to choose a set of three positive integers less than 100?

**Solution method**

Use the combinatorial formula: \(C_n^k = \frac{n!}{k! (n – k)!}\) = \(\frac{A^k_{n}}{k!}\ ), (\(0 k ≤ n\))

**Detailed explanation**

There are 99 positive integers less than 100.

+) Choose two positive integers less than 100, which is a convolution of 99 elements, so the number of ways to choose is: \(C_{99}^{2}= 4851\) ways.

+) Choose three positive integers less than 100, which is a convolution 3 of 99 elements, so the number of ways to choose is: \(C_{99}^{3}= 156849\) ways.

### Solve problem 8.9 page 70 Math textbook 10 Connecting knowledge volume 2

Ha has 5 green marbles and 7 red marbles. How many ways are there for Ha to choose exactly 2 marbles of different colors?

**Solution method**

Use the formula to find the union:

Number of ways to choose 1 green marble

Number of ways to choose 1 red marble

Use the multiplication rule

**Detailed explanation**

To select 2 balls of different colors, 1 blue and 1 red marble are selected.

The number of ways to choose a green marble is: \(C_{5}^{1}\) = 5 ways.

The number of ways to choose a red marble is: \(C_{7}^{1}\) = 7 ways.

So the number of ways to choose 2 marbles of different colors is: 5.7 = 35 ways.

### Solve problem 8.10, page 71 Math textbook 10 Connecting knowledge volume 2

A chess club has 10 boys and 7 girls. The coach wants to choose 4 friends to play chess.

a) How many ways are there to choose 4 boys?

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b) How many ways are there to choose 4 friends regardless of male or female?

c) How many ways are there to choose 4 friends, 2 boys and 2 girls?

**Solution method**

a) Choose 4 boys out of 10 boys as a handicap combination of 4 of 10 elements

b) Choose 4 people regardless of male or female from 17 you are a handicap combination of 4 of 17 elements

c) Number of ways to choose 2 boys out of 10 boys

Number of ways to choose 2 girls out of 7 girls

Use the multiplication rule

**Detailed explanation**

a) The number of ways to choose 4 boys is: \(C_{10}^{4}\) = 210 ways.

b) Number of ways to choose 4, regardless of male or female, is: \(C_{17}^{4}\) = 2380 ways.

c) Choose 2 boys out of 10, there are: \(C_{10}^{2}\) = 45 ways.

Choose 2 girls out of 7 girls, there are: \(C_{7}^{2}\) = 21 ways.

So the number of ways to choose 4 friends, 2 men and 2 women is: 45.21 = 945 ways.

### Solve problem 8.11 page 71 Math textbook 10 Connecting knowledge volume 2

How many natural numbers are there that are divisible by 5 that each have four distinct digits?

**Solution method**

– The natural number to be formed has the form \(\overline{abcd}\) and \(a, b,c, d\in A=\left \{ 0.1,2,3,4,5,6,7 ,8,9 \right \}, a\neq 0, a\neq b\neq c\neq d\).

+) Find the number of ways to choose the digit d so that \(\overline{abcd}\) is divisible by 5

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

+) Choose 3 numbers a, b, c and sort them from the set A\{d}

+) Find numbers of the form: \(\overline{0bc5}\)

+ Select b, c and sort from set A\{0; 5}

+ Infer the desired result

**Detailed explanation**

Call the 4-digit number to look up in the form: \(\overline{abcd}\) and \(a, b,c, d\in A=\left \{ 0,1,2,3,4,5, 6,7,8,9 \right \}, a\neq 0, a\neq b\neq c\neq d\).

For \(\overline{abcd}\) to be divisible by 5, d must be in the set {0; 5}.

Choose c has 2 ways,

Choose 3 numbers a, b, c and sort them from the set A\{d}, so the number of ways: \(A_{9}^{3}\) = 504 ways.

The number of ways to make it is: 504.2 = 1008 ways.

+ We find numbers of the form: \(\overline{0bc5}\),

Select b, c and sort from set A\{0; 5}, the number of ways is: \(A_{8}^{2}\) = 56 ways.

So the number of natural numbers divisible by 5 that have four different digits is: 1008 – 56 = 952 numbers.