Класс 11 RD Sharma Solutions – Глава 17 Комбинации – Упражнение 17.2 | Набор 3
Вопрос 23. Мы хотим выбрать 6 человек из 8, но если выбран человек А, то надо выбрать Б. Сколькими способами можно сделать выбор?
Решение:
Total number of persons = 8
Number of persons to be selected = 6
It is given that if the person A is chosen, then B must be chosen. Therefore, A and B are chosen together.
So, number of selections = 6C4
=
=
= 15
Also the number of selections in which A and B are not chosen = 7C6
=
= 7
Therefore, total number of ways = 15 + 7 = 22 ways.
Вопрос 24. Сколькими способами из 5 мальчиков и 4 девочек можно выбрать команду 3 мальчиков и 3 девочек?
Решение:
Total number of boys = 5
Total number of boys = 4
Number of boys to be selected = 3
Number of girls to be selected = 3
Hence, we have to choose 3 girls from 4 girls and 3 boys from 5 boys.
So, number of ways = 5C3 × 4C3
=
=
= 40
Therefore the required number of ways is 40.
Вопрос 25. Найдите количество способов выбрать 9 шаров из 6 красных, 5 белых и 5 синих шаров, если каждый выбор состоит из 3 шаров каждого цвета.
Решение:
We are given, 6 red balls, 5 white balls and 5 blue balls. We have to choose 3 balls of each colour.
Therefore, we can select 3 red balls from 6 red balls, 3 white balls from 5 white balls and 3 blue balls from 5 blue balls.
So, number of ways = 6C3 × 5C3 × 5C3
=
=
= 20×10×10
= 2000
Therefore, the required number of ways is 2000.
Вопрос 26. Определите количество комбинаций из 5 карт из колоды в 52 карты, если в каждой комбинации ровно один туз.
Решение:
Out of a deck of 52 cards, we have to choose 5 cards combinations where we want exactly one ace in each combination.
We know there are 4 aces in a deck of 52 cards. So, we have to choose 1 ace from 4 aces and selection of remaining cards will be from remaining 48 cards.
So, number of ways = (4C1 × 48C4)
=
=
= 778320
Therefore, number of ways of selecting 5 card combinations if there is exactly one ace in each combination is 778320.
Вопрос 27. Сколькими способами можно выбрать крикетную команду из одиннадцати игроков из 17, в которой только 5 человек могут играть в боулинг, если в каждой крикетной команде из 11 игроков должно быть ровно 4 боулера?
Решение:
Total number of players = 17
Total number of bowlers = 5
Number of bowlers to be selected = 4
As number of bowlers are 5, therefore number of batsmen = 17−5 = 12
To make a team of 11 players, we have to choose 7 batsmen from 12 batsmen and 4 bowlers from 5 bowlers.
So, number of ways = 12C7 × 5C4
=
=
= 3960
Therefore, number of ways of selecting the team is 3960.
Вопрос 28. В мешке 5 черных и 6 красных шаров. Определите, сколькими способами можно выбрать 2 черных и 3 красных шара.
Решение:
Total number of black balls = 5
Total number of red balls = 6
Now we have to choose 2 black balls from 5 black balls and 3 red balls from 6 red balls.
So, number of ways = 5C2 × 6C3
=
=
= 10×20
= 200
Therefore the required number of ways is 200.
Вопрос 29. Сколькими способами студент может выбрать программу из 5 курсов, если доступно 9 курсов и 2 определенных курса являются обязательными для каждого студента?
Решение:
Total number of courses = 9
Number of courses to be selected = 5
As 2 courses have been made compulsory, a student can only choose 3 courses (5−2) from remaining 7 courses (9−2).
So, number of ways = 7C3
=
=
= 35
Therefore, the required number of ways is 35.
Вопрос 30. Необходимо сформировать комиссию 7 человек из 9 мальчиков и 4 девочек. Сколькими способами это можно сделать, если комитет состоит из
(i) ровно 3 девушки?
(ii) не менее 3 девочек?
(iii) не более 3 девочек?
Решение:
Общее количество мальчиков = 9
Общее количество девушек = 4
Количество человек, которые будут выбраны = 7
(i) ровно 3 девушки
If we want exactly 3 girls in the committee, we have to choose 3 girls from 4 girls and 4 boys from 9 boys.
So, number of ways = 9C4 × 4C3
=
=
= 9×8×7
= 504
Therefore, the number of ways of selecting a committee in which exactly 3 girls are required is 504.
(ii) не менее 3 девочек?
As the required number of girls can be 3 or more than 3, we can choose 3 or 4 girls and adjust the number of boys according to it.
So, number of ways = (9C4 × 4C3) + (9C3 × 4C4)
=
=
= 504 + 84
= 588
Therefore, the number of ways of selecting a committee in which at least 3 girls are required is 588.
(iii) не более 3 девочек.
As the required number of girls can be 3 or less than 3, we can choose 3 or 2 or 1 or 0 number of girls and adjust the number of boys according to it.
So, number of ways = (4C0 × 9C7) + (4C1 × 9C6) + (4C2 × 9C5) + (4C3 × 9C4)
=
=
= 36 + 336 + 756+ 504
= 1632
Therefore, the number of ways of selecting a committee in which at most 3 girls are required is 1632.
Вопрос 31. В экзаменационном листе вопросник состоит из 12 вопросов, разделенных на две части Часть I и Часть II, содержащие 5 и 7 вопросов соответственно. Всего учащийся должен ответить на 8 вопросов, выбрав не менее 3 из каждой части. Сколькими способами учащийся может выбрать вопросы?
Решение:
Number of questions in Part I = 5
Number of questions in Part II = 7
Number of questions to be attempted = 8
Now a student has to attempt at least 3 questions from each part and 8 in total.
So, number of ways = (5C3 × 7C5) + (5C4 × 7C4) + (5C5 × 7C3)
=
=
= 210 + 175 + 35
= 420
Therefore, the number of ways a student can select the questions is 420.
Вопрос 32. Параллелограмм разрезан двумя наборами m прямых, параллельных его сторонам. Найдите количество образовавшихся параллелограммов.
Решение:
A parallelogram has 2 sets of parallel lines. Now m lines are parallel to these sets. Therefore, total number of lines for one set is (m+2) lines.
Therefore the total number of parallelograms formed = m+2C2 × m+2C2
=
= (m+2)2 (m+1)2/4
Therefore, the number of parallelogram formed is (m+2)2 (m+1)2/4.
Вопрос 33. Из 18 точек плоскости никакие три не лежат на одной прямой, кроме пяти точек, лежащих на одной прямой. Как много
(i) прямые линии могут быть образованы путем их соединения?
Решение:
Given there are 18 points in a plane where 5 points are collinear.
Now we know, number of lines formed will be the difference between total number of lines formed by all 18 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 5 collinear points.
So, number of ways = 18C2 – 5C2 + 1
=
=
= 144
Therefore number of straight lines formed by joining 18 points is 144.
(ii) треугольники можно образовать, соединив их?
Решение:
Now three points are required to form a triangle.
Number of points that can be used to make triangles = 18−5 = 13
So, number of triangles = 13C3
=
=
= 13×22
= 286
Therefore number of triangles formed by joining 18 points is 286.
