Класс 11 RD Sharma Solutions – Глава 17 Комбинации – Упражнение 17.2 | Набор 2
Вопрос 12. На экзамене студент должен ответить на 4 вопроса из 5 вопросов; однако вопросы 1 и 2 являются обязательными. Определите количество способов, которыми учащийся может сделать выбор.
Решение:
Total number of questions = 5
Total number of questions to be answered = 4
As 2 questions are compulsory to answer, student can choose only 2 questions (4−2) out of the remaining 3 questions (5−2).
So, number of ways = 3C2
=
=
= 3
Therefore, the number of ways answering the questions is 3.
Вопрос 13. От кандидата требуется ответить на 7 вопросов из 12 вопросов, которые разбиты на две группы по 6 вопросов в каждой. Ему не разрешается задавать более 5 вопросов из любой группы. Сколькими способами он может выбрать 7 вопросов?
Решение:
Total number of questions = 12
Total number of questions to be answered = 7
Total number of questions in each set = 6
Now a student can’t attempt 5 questions from either group but has to answer 7 questions in total.
So, number of ways = (6C5 × 6C2) + (6C4 × 6C3) + (6C3 × 6C4) + (6C2 × 6C5)
=
=
= (6×15) + (15×20) + (20×15) + (15×6)
= 90 + 300 + 300 + 90
= 780
Therefore, the number of ways of answering 7 questions is 780.
Вопрос 14. На плоскости 10 точек, из которых 4 коллинеарны. Сколько различных прямых можно провести, соединив эти точки.
Решение:
Total number of points = 10
Number of collinear points = 4
Now we know, number of lines formed will be the difference between total number of lines formed by all 10 points and number of lines formed by collinear points added with 1.
Here, we add 1 because only 1 line can be formed by the given four collinear points.
So, number of ways = 10C2 – 4C2 + 1
=
=
= 45 – 6 + 1
= 40
Therefore, the total number of ways of drawing different lines is 40.
Вопрос 15. Найдите количество диагоналей
(и) шестиугольник
Решение:
A hexagon has 6 angular points. By joining any two angular points we get a line which is either a side or a diagonal.
So number of lines formed = 6C2
=
=
= 3×5
= 15
We know number of sides of hexagon is 6.
So, number of diagonals = 15 – 6 = 9
Therefore, the total number of diagonals of a hexagon is 9.
(ii) многоугольник из 16 сторон
Решение:
A polygon of 16 sides has 16 angular points. By joining any two angular points we get a line which is either a side or a diagonal.
So number of lines formed = 16C2
=
=
= 8×15
= 120
Number of sides of given polygon = 16
So, number of diagonals = 120 – 16 = 104
Therefore, the total number of diagonals of a hexagon is 104.
Вопрос 16. Сколько треугольников можно получить, соединив 12 точек, пять из которых лежат на одной прямой?
Решение:
We know that 3 points are required to draw a triangle.
Since 5 out of 12 points are collinear, so number of triangles that can be formed would be the difference between the number of triangles formed by all 12 points and the number of triangles formed by collinear points.
So, number of triangle= 12C3 – 5C3
=
=
= (2×11×10) – (5×2)
= 220 – 10
= 210
Therefore, the total number of triangles that can be formed is 210.
Вопрос 17. Сколькими способами можно составить комитет из 5 человек из 6 мужчин и 4 женщин, если обязательно должна быть выбрана хотя бы одна женщина?
Решение:
Total number of men = 6
Total number of women = 4
Total number of persons to be selected = 5.
Now it is given that we must choose at least one woman. It means we can choose from 1 to all 4 women in our committee at a time. Number of men will change according to it.
So, number of ways = (4C1 × 6C4) + (4C2 × 6C3) + (4C3 × 6C3) + (4C4 × 6C1)
=
=
= 60 + 120 + 60 + 6
= 246
Therefore, the number of ways of selection is 246.
Вопрос 18. В деревне 87 семей, из них 52 семьи имеют не более 2 детей. В рамках программы развития сельских районов необходимо выбрать для помощи 20 семей, из которых 18 семей должны иметь не более 2 детей. Сколькими способами можно сделать выбор?
Решение:
Its given that 52 families have at most 2 children out of 87. Therefore the remaining 35 families have exactly 2 children.
Now to choose any 20 families of which 18 families must have at most 2 children can be done in 3 ways. Either we can choose all the 20 families out of those 52 who have at most children, or 19 out of 52 and remaining 1 out of 35, or 18 families from 52 and remaining 2 out of 35.
So, total number of ways = (52C18 × 35C2) + (52C19 × 35C1) + (52C20 × 35C0)
Вопрос 19. Группа состоит из 4 девочек и 7 мальчиков. Сколькими способами можно выбрать команду из 5 человек, если в команде
(и) нет девушек
(ii) по крайней мере один мальчик и девочка
(iii) не менее 3 девочек
Решение:
Количество девушек = 4
Количество мальчиков = 7
Количество участников, которые будут выбраны = 5
(и) нет девушек
As it is given that the team cannot have a girl, we have to choose 5 members out of 7 boys.
So, number of ways = 7C5
=
=
= 21
Therefore, the number of ways of selection such that team has no girls is 21.
(ii) по крайней мере один мальчик и девочка
To select a team which consists of at least one boy and girl, we can choose from 1 to all the 4 girls at a time. Number of boys will change according to it.
So, number of ways = (7C1 × 4C4) + (7C2 × 4C3) + (7C3 × 4C3) + (7C4 × 4C1)
=
=
= 7+84+210+140
= 441
Therefore, the number of ways of selection such that team has at least one boy and girl is 441.
(iii) не менее 3 девочек
To select a team which consists of at least 3 girls, we can choose either 3 or 4 girls at a time. Number of boys will change according to it.
So, number of ways = (7C2 × 4C3) + (7C1 × 4C4)
=
=
= 84 + 7
= 91
Therefore, the number of ways of selection such that team has at least 3 girls is 91.
Вопрос 20. Из группы 2 мужчин и 3 женщин сформировать комиссию в составе 3 человек. Сколькими способами это можно сделать? Сколько из этих комитетов будет состоять из 1 мужчины и 2 женщин?
Решение:
Total number of men = 2
Total number of women = 3
So total number of persons = 2+3 = 5
Number of persons to be selected = 3
It is given that a committee of 3 persons has to be formed. So we have to choose 3 persons from 5 persons.
So, number of ways = 5C3
=
=
= 10
Also we have to find the number of committees consisting of 1 man and 2 women. So we have to choose 1 man out of 2 men and 2 women out of 3 women.
So, number of ways = 2C1 × 3C2
=
= 6
Вопрос 21. Найдите количество
(i) диагонали, образованные десятиугольником.
Решение:
A decagon has 10 sides. By joining any two angular points we get a line which is either a side or a diagonal.
So number of lines formed = 10C2
=
=
= 45
Number of sides = 10
So, number of diagonals = 45−10 = 35.
Therefore, number of diagonals formed in a decagon is 35.
(ii) треугольники, образованные десятиугольником.
Решение:
A decagon has 10 sides. By joining any 3 angular points we get a triangle.
So number of lines formed = 10C3
=
=
=
= 120
Therefore, number of triangles formed in a decagon is 120.
Вопрос 22. Определите количество комбинаций из 5 карт в колоде из 52 карт, если хотя бы одна из 5 карт должна быть королем?
Решение:
Out of a deck of 52 cards, we have to choose 5 cards combinations where at least one of the 5 cards has to be a king.
We know there are 4 kings in a deck of 52 cards. So, we can choose from 1 to all the four kings at a time and selection of remaining cards will change accordingly.
So, number of ways = (4C1 × 48C4) + (4C2 × 48C3) + (4C3 × 48C2) + (4C4 × 48C1)
=
=
= 778320 + 103776 + 4512 + 48
= 886656
Therefore, number of ways of selecting 5 card combinations if at least one of the 5 cards has to be a king is 886656.
