Количество перестановок букв в слове экзамен равно

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Источник: Демонстрационная версия ЕНТ по математической грамотности 2022 года, вариант 2. Отредактировано редакцией Решу ЕНТ в формат актуальной демоверсии

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›Презентации›Презентация по математике (алгебра и начала математического анализа) на тему «Перестановки»

Презентация по математике на тему «Перестановки»

Question 1: How many different words, each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants.

Solution:

Given, word contains 2 vowels and 3 consonants.

So, we need to choose 2 vowels out of 5 vowels and 3 consonants out of 17 consonants.

This can be done in  ways.

Also, we need number of different words, we can arrange 5 letter word in 5! ways.

Therefore, Total number of ways = 

⇒ 6800 × 120 = 816000

Question 2: There are 10 persons named P₁,P₂,P₃….P₁₀ out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P₁ must occur whereas P₄ and P₅ do not occur. Find the number of such possible arrangements.

Solution: 

Given, we need to arrange 5 persons out of 10 persons such that in each arrangements P₁ must occur whereas P₄ and P₅ do not occur.

Here, we Should choose P₁ every time, so now, we choose 4 persons out of 9 persons.

From that 9 persons, we don’t need to choose P₄ and P₅, so This can be done in  ways.

Therefore, number of selections = 

And, 5 persons can be arranged in 5! ways. so,

Total number of ways = 

⇒ 4200.

Therefore, number of such possible arrangements is 4200.

Question 3: How many words, with or without meaning can be formed from the letters of the word “MONDAY” assuming that no letter is repeated. if

(i) 4 letters are used at a time.

Solution:

Given, six letter word “MONDAY”.

4 letters are used at a time out of 6 letters. this can be done in  ways.

These four letters can be arranged in 4! ways.

Therefore, total number of ways = 

⇒ 360.

(ii) All letters are used at a time.

Solution:

Given, six letter word “MONDAY”.

6 letters are used at a time out of 6 letters. this can be done in  ways.

These 6 letters can be arranged in 6! ways.

Therefore, Total number of ways = 

⇒ 1 × 720 = 720

(iii) All letters are used but first letter is a vowel.

Solution:

Given, six letter word “MONDAY”.

All letters are used at a time but first letter is Vowel,

number of vowels in word “MONDAY” is 2. So first we choose one vowel from these two in  ways.

The remaining 5 letters out of 5 letters. This can be done in  ways.

Number of arrangements can be done for these 5 letters are 5!.

Therefore, total number of ways = 

⇒ 2times1times120=240.

Question 4: Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.

Solution:

Given, Number of permutations of n distinct things taking r together and 3 particular things are already selected.

So, now number of ways of choosing (r – 3) things from remaining (n – 3) things is,

 This can be done in  ways.

We need to find the number of permutations, there are total (r – 2) things considering 3 particular things as single thing.

These can be arranged in (r – 2)! ways.

Internally, 3 particular things can be arranged in 3! ways.

Therefore, Total number of permutations = 

Question 5: How many words of each 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?

Solution:

Given word is “INVOLUTE”.

Number of vowels and consonants in the word are 4 and 4 respectively.

Number of ways in choosing 3 vowels out of 4 and 2 consonants out of 4 is 

These five letters can be arranged in 5! ways.

Therefore, Total number of words formed = 

⇒ 4 × 6 × 120 = 2880

Question 6: Find the number of permutations of n different things taken r at a time such that two specified things occur together?

Solution: 

Given, number of permutations of n different things taken r at a time and two specified things occur together.

So, we now choose (r – 2) things from the remaining (n – 2) things.

This can be done in  ways.

We need to find the number of permutations, there are total (r – 1) things considering 2 specific things as single thing.

These can be arranged in (r – 1)! ways.

Internally, 2 particular things can be arranged in 2! ways.

Therefore, Total number of permutations = 

⇒ 

⇒ 

Question 7: Find the number of ways in which: (a) a selection (b) an arrangement, of four letters can be made from the letters of the word ‘PROPORTION’.

Solution: 

Given word is ‘PROPORTION’

there are 10 letters in this word and mainly ‘OOO’, ‘PP’, ‘RR’, ‘I’, ‘T’, ‘N’. 

(a) Here we need to select 4 letters out of 10 letters. but we need to consider some cases.

1. 3 alike letters and 1 distinct letter
2. 2 alike letters and 2 distinct letters
3. 2 alike letters of I kind and 2 alike letters of other kind.
4. all distinct letters.

lets consider,

case-1: 3 alike and 1 distinct 

here we find that only one 3 alike letter in the word (‘OOO’).

So, choosing 1 distinct letter from remaining 5 distinct letters is 

⇒ 5 ways.

case-2: 2 alike letters and 2 distinct letters.

Here, we find that 3 chars have 2 alike letters , number of ways in choosing 1 letters out of 3 is  

And number of ways of choosing 2 distinct letters from remaining 5 letters is 

⇒   ways.

case-3 : 2 alike letters of I kind and 2 alike letters of other kind.

Here, we find 3 chars have 2 alike letters, number of ways in choosing 2 letters out of 3 is 

⇒ 3 ways

Case-4: all distinct letters.

Here, we have 6 different letters, number of ways in choosing 4 letters out of 6 is 

⇒ 15 ways.

Therefore, Total number of ways is sum of number of ways in all cases.

⇒ 5+30+3+15=53 ways.

(b) Here we need to arrange 4 letters out of 10 letters, here cases will be same as selection, but we arrange the given letters in every case.

case-1: Number of ways of selecting 3 alike and 1 distinct letters is 5.

Number of ways in arranging is similar to arranging the n people in n places where r places are same = 

similarly, arrangement of 4 letters where 3 are alike is 

total number of ways are 

case-2: Number of ways of selecting 2 alike letters and 2 distinct letters is 30 ways.

number ways of arranging them is 

Total number of ways are 

case-3: Number of ways of selecting 2 letters alike and 2 letters alike is 3 ways.

Number of ways in arranging them is 

Total number of ways are 

Case-4: Number of ways of selecting 4 distinct letters is 15 ways.

number of ways of arranging them is 4!

Total number of ways are 15 × 4! = 360

considering all cases, Total number of ways are sum of all number of ways of all cases

⇒ 20+360+18+360 = 758 ways.

Question 8: How many words can be formed by taking 4 letters at a time from the letters of word ‘MORADABAD’?

Solution: 

Given word is ‘MORADABAD’

There are 10 letters in this word and mainly ‘AAA’, ‘DD’, ‘M’, ‘R’, ‘B’, ‘O’.

(a) Here we need to select 4 letters out of 10 letters. but we need to consider some cases.

1. 3 alike letters and 1 distinct letter
2. 2 alike letters and 2 distinct letters
3. 2 alike letters of I kind and 2 alike letters of other kind.
4. all distinct letters.

lets consider,

case-1: 3 alike and 1 distinct

here we find that only one 3 alike letter in the word (‘AAA’).

So, choosing 1 distinct letter from remaining 5 distinct letters is 

Number of ways of arranging them is 

⇒  ways.

case-2: 2 alike letters and 2 distinct letters.

Here, we find that 2 chars have 2 alike letters , number of ways in choosing 1 letters out of 2 is 

And number of ways of choosing 2 distinct letters from remaining 5 letters is 

Number of ways of arranging them is 

⇒  ways.

case-3 : 2 alike letters of I kind and 2 alike letters of other kind.

Here, we find 2 chars have 2 alike letters, number of ways in choosing 2 letters out of 2 is 

Number of ways of arranging them is 

⇒  ways

Case-4: all distinct letters.

Here, we have 6 different letters, number of ways in choosing 4 letters out of 6 is 

Number of ways of arranging them is 4!

⇒  ways.

Therefore, Total number of ways is sum of number of ways in all cases.

⇒ 20+240+6+360 = 626 ways.

Question 9: A businessman hosts a dinner to 21 guests. He is having 2 round tables which can accommodate 15 and 6 persons each. In how many ways can he arrange the guests?

Solution:

Given, A business man hosts dinners to 21 guests, where 2 round tables accommodate 16 and 6 persons.

So, choosing 15 guests out of 21 to accommodate in one table in  ways.

Those 15 guests can be arranged in themselves in (15 – 1)! ways. because it is a round table. 

So, we need to keep a guest fixed and arrange remaining 14 guests = (15 – 1)! = 14!.

After accommodating 15 guests in one table, accommodating remaining 6 guests out of 6 in another table in  ways.

Those 6 guests can be arranged themselves in (6 – 1)! = 5! ways.

Total number of ways = 

Question 10: Find the number of combinations and permutations of 4 letters taken from the word ‘EXAMINATION’.

Solution:

Given word is ‘EXAMINATION’

There are 10 letters in this word and mainly ‘AA’, ‘NN’, ‘II’, ‘E’, ‘X’, ‘O’,’M’,’T’.

(a) Here we need to select 4 letters out of 10 letters. but we need to consider some cases.

1. 2 alike letters and 2 distinct letters
2. 2 alike letters of I kind and 2 alike letters of other kind.
3. all distinct letters.

lets consider,

case-1: 2 alike letters and 2 distinct letters.

Here, we find that 3 chars have 2 alike letters, number of ways in choosing 1 letters out of 3 is 

And number of ways of choosing 2 distinct letters from remaining 7 letters is 

Number of ways of arranging them is 

⇒  ways.

case-2 : 2 alike letters of I kind and 2 alike letters of other kind.

Here, we find 3 chars have 2 alike letters, number of ways in choosing 2 letters out of 3 is 

Number of ways of arranging them is 

⇒  ways

Case-3: all distinct letters.

Here, we have 6 different letters, number of ways in choosing 4 letters out of 8 is 

Number of ways of arranging them is 4!

⇒  ways.

Therefore, Total number of ways is sum of number of ways in all cases.

⇒ 756+18+1680 = 2454 ways.

Question 11: A tea party is arranged for 16 persons along two sides of a long table with 8 chairs on each side. Four persons wish to sit on one particular side and two on another side. In how many ways can they be seated?

Solution: 

Given, A tea party is arranged 16 persons along two sides of long table with 8 chairs on each side.

Let two sides be side A, side B.

Also, 4 persons wish to sit on side A, and 2 persons on side B.

Remaining seats in side A and side B are 4, 6 respectively.

Now, let we choose 4 persons outs of remaining 10 on side A. 

This can be done in  ways.

Also, choosing 6 persons out of remaining 6 on side B.

This can be done in  ways.

Now, 8 persons on each side can be arranged in 8! ways.

Therefore, total number of ways = 

⇒ 

Question 1: How many different words, each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants.

Solution:

Given, word contains 2 vowels and 3 consonants.

So, we need to choose 2 vowels out of 5 vowels and 3 consonants out of 17 consonants.

This can be done in  ways.

Also, we need number of different words, we can arrange 5 letter word in 5! ways.

Therefore, Total number of ways = 

⇒ 6800 × 120 = 816000

Question 2: There are 10 persons named P₁,P₂,P₃….P₁₀ out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P₁ must occur whereas P₄ and P₅ do not occur. Find the number of such possible arrangements.

Solution: 

Given, we need to arrange 5 persons out of 10 persons such that in each arrangements P₁ must occur whereas P₄ and P₅ do not occur.

Here, we Should choose P₁ every time, so now, we choose 4 persons out of 9 persons.

From that 9 persons, we don’t need to choose P₄ and P₅, so This can be done in  ways.

Therefore, number of selections = 

And, 5 persons can be arranged in 5! ways. so,

Total number of ways = 

⇒ 4200.

Therefore, number of such possible arrangements is 4200.

Question 3: How many words, with or without meaning can be formed from the letters of the word “MONDAY” assuming that no letter is repeated. if

(i) 4 letters are used at a time.

Solution:

Given, six letter word “MONDAY”.

4 letters are used at a time out of 6 letters. this can be done in  ways.

These four letters can be arranged in 4! ways.

Therefore, total number of ways = 

⇒ 360.

(ii) All letters are used at a time.

Solution:

Given, six letter word “MONDAY”.

6 letters are used at a time out of 6 letters. this can be done in  ways.

These 6 letters can be arranged in 6! ways.

Therefore, Total number of ways = 

⇒ 1 × 720 = 720

(iii) All letters are used but first letter is a vowel.

Solution:

Given, six letter word “MONDAY”.

All letters are used at a time but first letter is Vowel,

number of vowels in word “MONDAY” is 2. So first we choose one vowel from these two in  ways.

The remaining 5 letters out of 5 letters. This can be done in  ways.

Number of arrangements can be done for these 5 letters are 5!.

Therefore, total number of ways = 

⇒ 2times1times120=240.

Question 4: Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.

Solution:

Given, Number of permutations of n distinct things taking r together and 3 particular things are already selected.

So, now number of ways of choosing (r – 3) things from remaining (n – 3) things is,

 This can be done in  ways.

We need to find the number of permutations, there are total (r – 2) things considering 3 particular things as single thing.

These can be arranged in (r – 2)! ways.

Internally, 3 particular things can be arranged in 3! ways.

Therefore, Total number of permutations = 

Question 5: How many words of each 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?

Solution:

Given word is “INVOLUTE”.

Number of vowels and consonants in the word are 4 and 4 respectively.

Number of ways in choosing 3 vowels out of 4 and 2 consonants out of 4 is 

These five letters can be arranged in 5! ways.

Therefore, Total number of words formed = 

⇒ 4 × 6 × 120 = 2880

Question 6: Find the number of permutations of n different things taken r at a time such that two specified things occur together?

Solution: 

Given, number of permutations of n different things taken r at a time and two specified things occur together.

So, we now choose (r – 2) things from the remaining (n – 2) things.

This can be done in  ways.

We need to find the number of permutations, there are total (r – 1) things considering 2 specific things as single thing.

These can be arranged in (r – 1)! ways.

Internally, 2 particular things can be arranged in 2! ways.

Therefore, Total number of permutations = 

⇒ 

⇒ 

Question 7: Find the number of ways in which: (a) a selection (b) an arrangement, of four letters can be made from the letters of the word ‘PROPORTION’.

Solution: 

Given word is ‘PROPORTION’

there are 10 letters in this word and mainly ‘OOO’, ‘PP’, ‘RR’, ‘I’, ‘T’, ‘N’. 

(a) Here we need to select 4 letters out of 10 letters. but we need to consider some cases.

1. 3 alike letters and 1 distinct letter
2. 2 alike letters and 2 distinct letters
3. 2 alike letters of I kind and 2 alike letters of other kind.
4. all distinct letters.

lets consider,

case-1: 3 alike and 1 distinct 

here we find that only one 3 alike letter in the word (‘OOO’).

So, choosing 1 distinct letter from remaining 5 distinct letters is 

⇒ 5 ways.

case-2: 2 alike letters and 2 distinct letters.

Here, we find that 3 chars have 2 alike letters , number of ways in choosing 1 letters out of 3 is  

And number of ways of choosing 2 distinct letters from remaining 5 letters is 

⇒   ways.

case-3 : 2 alike letters of I kind and 2 alike letters of other kind.

Here, we find 3 chars have 2 alike letters, number of ways in choosing 2 letters out of 3 is 

⇒ 3 ways

Case-4: all distinct letters.

Here, we have 6 different letters, number of ways in choosing 4 letters out of 6 is 

⇒ 15 ways.

Therefore, Total number of ways is sum of number of ways in all cases.

⇒ 5+30+3+15=53 ways.

(b) Here we need to arrange 4 letters out of 10 letters, here cases will be same as selection, but we arrange the given letters in every case.

case-1: Number of ways of selecting 3 alike and 1 distinct letters is 5.

Number of ways in arranging is similar to arranging the n people in n places where r places are same = 

similarly, arrangement of 4 letters where 3 are alike is 

total number of ways are 

case-2: Number of ways of selecting 2 alike letters and 2 distinct letters is 30 ways.

number ways of arranging them is 

Total number of ways are 

case-3: Number of ways of selecting 2 letters alike and 2 letters alike is 3 ways.

Number of ways in arranging them is 

Total number of ways are 

Case-4: Number of ways of selecting 4 distinct letters is 15 ways.

number of ways of arranging them is 4!

Total number of ways are 15 × 4! = 360

considering all cases, Total number of ways are sum of all number of ways of all cases

⇒ 20+360+18+360 = 758 ways.

Question 8: How many words can be formed by taking 4 letters at a time from the letters of word ‘MORADABAD’?

Solution: 

Given word is ‘MORADABAD’

There are 10 letters in this word and mainly ‘AAA’, ‘DD’, ‘M’, ‘R’, ‘B’, ‘O’.

(a) Here we need to select 4 letters out of 10 letters. but we need to consider some cases.

1. 3 alike letters and 1 distinct letter
2. 2 alike letters and 2 distinct letters
3. 2 alike letters of I kind and 2 alike letters of other kind.
4. all distinct letters.

lets consider,

case-1: 3 alike and 1 distinct

here we find that only one 3 alike letter in the word (‘AAA’).

So, choosing 1 distinct letter from remaining 5 distinct letters is 

Number of ways of arranging them is 

⇒  ways.

case-2: 2 alike letters and 2 distinct letters.

Here, we find that 2 chars have 2 alike letters , number of ways in choosing 1 letters out of 2 is 

And number of ways of choosing 2 distinct letters from remaining 5 letters is 

Number of ways of arranging them is 

⇒  ways.

case-3 : 2 alike letters of I kind and 2 alike letters of other kind.

Here, we find 2 chars have 2 alike letters, number of ways in choosing 2 letters out of 2 is 

Number of ways of arranging them is 

⇒  ways

Case-4: all distinct letters.

Here, we have 6 different letters, number of ways in choosing 4 letters out of 6 is 

Number of ways of arranging them is 4!

⇒  ways.

Therefore, Total number of ways is sum of number of ways in all cases.

⇒ 20+240+6+360 = 626 ways.

Question 9: A businessman hosts a dinner to 21 guests. He is having 2 round tables which can accommodate 15 and 6 persons each. In how many ways can he arrange the guests?

Solution:

Given, A business man hosts dinners to 21 guests, where 2 round tables accommodate 16 and 6 persons.

So, choosing 15 guests out of 21 to accommodate in one table in  ways.

Those 15 guests can be arranged in themselves in (15 – 1)! ways. because it is a round table. 

So, we need to keep a guest fixed and arrange remaining 14 guests = (15 – 1)! = 14!.

After accommodating 15 guests in one table, accommodating remaining 6 guests out of 6 in another table in  ways.

Those 6 guests can be arranged themselves in (6 – 1)! = 5! ways.

Total number of ways = 

Question 10: Find the number of combinations and permutations of 4 letters taken from the word ‘EXAMINATION’.

Solution:

Given word is ‘EXAMINATION’

There are 10 letters in this word and mainly ‘AA’, ‘NN’, ‘II’, ‘E’, ‘X’, ‘O’,’M’,’T’.

(a) Here we need to select 4 letters out of 10 letters. but we need to consider some cases.

1. 2 alike letters and 2 distinct letters
2. 2 alike letters of I kind and 2 alike letters of other kind.
3. all distinct letters.

lets consider,

case-1: 2 alike letters and 2 distinct letters.

Here, we find that 3 chars have 2 alike letters, number of ways in choosing 1 letters out of 3 is 

And number of ways of choosing 2 distinct letters from remaining 7 letters is 

Number of ways of arranging them is 

⇒  ways.

case-2 : 2 alike letters of I kind and 2 alike letters of other kind.

Here, we find 3 chars have 2 alike letters, number of ways in choosing 2 letters out of 3 is 

Number of ways of arranging them is 

⇒  ways

Case-3: all distinct letters.

Here, we have 6 different letters, number of ways in choosing 4 letters out of 8 is 

Number of ways of arranging them is 4!

⇒  ways.

Therefore, Total number of ways is sum of number of ways in all cases.

⇒ 756+18+1680 = 2454 ways.

Question 11: A tea party is arranged for 16 persons along two sides of a long table with 8 chairs on each side. Four persons wish to sit on one particular side and two on another side. In how many ways can they be seated?

Solution: 

Given, A tea party is arranged 16 persons along two sides of long table with 8 chairs on each side.

Let two sides be side A, side B.

Also, 4 persons wish to sit on side A, and 2 persons on side B.

Remaining seats in side A and side B are 4, 6 respectively.

Now, let we choose 4 persons outs of remaining 10 on side A. 

This can be done in  ways.

Also, choosing 6 persons out of remaining 6 on side B.

This can be done in  ways.

Now, 8 persons on each side can be arranged in 8! ways.

Therefore, total number of ways = 

⇒ 

1


Перестановки

2


Перестановки Определение 1 Перестановкой из n элементов называется всякий способ нумерации этих элементов Пример 1 Дано множество. Составить все перестановки этого множества. Решение.

3


Число перестановок Теорема 1. Число всех различных перестановок из n элементов равно n! Замечание. Например, Считают, что 0!=1 читается «n факториал» и вычисляется по формуле

4


Число перестановок Доказательство теоремы 1. Любую перестановку из n элементов можно получить с помощью n действий: 1)выбор первого элемента n различными способами, 2)выбор второго элемента из оставшихся (n-1) элементов, т.е. (n-1) способом, 3)выбор третьего элемента (n-2) способами, …… n) выбор n-го элемента 1 способом. По правилу умножения число всех способов выполнения действий, т.е. число перестановок, равно Теорема доказана.

5


Перестановки Число всех перестановок обозначается Итак, Пример В команде 6 человек. Сколькими способами они могут построиться для приветствия? Решение Число способов построения равно числу перестановок 6 элементов, т.е.

6


Перестановки с повторениями Теорема 2 Число перестановок n – элементов, в котором есть одинаковые элементы, а именно элементов i –того типа ( ) вычисляется по формуле где Доказательство. Так как перестановки между одинаковыми элементами не изменяют вид перестановки в целом, количество перестановок всех элементов множества нужно разделить на число перестановок одинаковых элементов.

7


Пример Задача: Сколько слов можно составить, переставив буквы в слове «экзамен», а в слове «математика»? Решение: В слове «экзамен» все буквы различны, поэтому используем формулу для числа перестановок без повторений В слове «математика» 3 буквы «а», 2 буквы «м», 2 буквы «т», поэтому число перестановок всех букв разделим на число перестановок повторяющихся букв:

8


Размещения

9


Размещения Определение 1 Размещением из n элементов по k называется всякая перестановка из k элементов, выбранных каким-либо способом из данных n. Пример Дано множество. Составим все 2- размещения этого множества.

10


Число размещений Теорема 1 Число всех размещений из n элементов по k вычисляется по формуле Доказательство. Каждое размещение можно получить с помощью k действий: 1) выбор первого элемента n способами; 2) выбор второго элемента (n-1) способами; и т. д. k) выбор k –го элемента (n-(k-1))=(n-k+1) способами. По правилу умножения число всех размещений будет n(n-1)(n-2)…(n-k+1). Теорема доказана.

11


Число размещений Замечание. Формулу для числа размещений можно записать в виде Действительно

12


Пример Абонент забыл последние 3 цифры номера телефона. Какое максимальное число номеров ему нужно перебрать, если он вспомнил, что эти последние цифры разные? Решение. Задача сводится к поиску различных перестановок 3 элементов из 10 ( так как всего цифр 10). Применим формулу для числа перестановок.

13


Размещения с повторениями Определение 2 Размещением с повторением из n элементов по k называется всякая перестановка из k элементов, выбранных каким-либо способом из данных n элементов возможно с повторениями. Пример Дано множество Составим 2- размещения с повторениями:

14


Число размещений с повторениями Теорема 2. Число k- размещений с повторениями из n элементов вычисляется по формуле Доказательство. Каждый элемент размещения можно выбрать n способами. По правилу умножения число всех размещений с повторениями равно

15


Пример Сколько существует номеров машин? Решение. Считаем, что в трех буквах номера машины не используются буквы «й», «ы», «ь», «ъ», тогда число перестановок букв равно. Число перестановок цифр равно. По правилу умножения получим число номеров машин

16


Решение задач

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Задачи 1)Сколькими способами можно составить список из 8 учеников, если нет полного совпадения ФИО ? Решение Задача сводится к подсчету числа перестановок ФИО.

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Задачи 2)Сколькими способами можно составить список 8 учеников, так, чтобы два указанных ученика располагались рядом? Решение Можно считать двоих указанных учеников за один объект и считать число перестановок уже 7 объектов, т.е. Так как этих двоих можно переставлять местами друг с другом, необходимо умножить результат на 2!

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Задачи 3) Сколькими способами можно разделить 11 спортсменов на 3 группы по 4, 5 и 2 человека соответственно? Решение. Сделаем карточки: четыре карточки с номером 1, пять карточек с номером 2 и две карточки с номером 3. Будем раздавать эти карточки с номерами групп спортсменам, и каждый способ раздачи будет соответствовать разбиению спортсменов на группы. Таким образом нам необходимо посчитать число перестановок 11 карточек, среди которых четыре карточки с одинаковым номером 1, пять карточек с номером 2 и две карточки с номером 3.

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Задачи 4) Сколькими способами можно вызвать по очереди к доске 4 учеников из 7? Решение. Задача сводится к подсчету числа размещений из 7 элементов по 4

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Задачи 5)Сколько существует четырехзначных чисел, у которых все цифры различны? Решение. В разряде единиц тысяч не может быть нуля, т.е возможны 9 вариантов цифры. В остальных трех разрядах не может быть цифры, стоящей в разряде единиц тысяч (так как все цифры должны быть различны), поэтому число вариантов вычислим по формуле размещений без повторений из 9 по 3 По правилу умножения получим

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Задачи 6)Сколько существует двоичных чисел, длина которых не превосходит 10? Решение. Задача сводится к подсчету числа размещений с повторениями из двух элементов по 10

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Задачи 7)В лифт 9 этажного дома зашли 7 человек. Сколькими способами они могут распределиться по этажам дома? Решение. Очевидно, что на первом этаже никому не надо выходить. Каждый из 7 человек может выбрать любой из 8 этажей, поэтому по правилу умножения получим Можно так же применить формулу для числа размещений с повторениями из 8 (этажей) по 7(на каждого человека по одному этажу)

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Задачи 8)Сколько чисел, меньше можно написать с помощью цифр 2,7,0? Решение. Так как среди цифр есть 0, то, например запись 0227 соответствует числу 227, запись 0072 соответствует числу 72, а запись 0007 соответствует числу 7. Таким образом, задачу можно решить, используя формулу числа размещений с повторениями