Класс 11 RD Sharma Solutions – Глава 17 Комбинации – Упражнение 17.2 | Набор 1
Вопрос 1. Из группы из 15 игроков в крикет необходимо выбрать команду из 11 игроков. Сколькими способами это можно сделать?
Решение:
We are given,
Total number of players = 15
Number of players to be chosen = 11
So, number of ways = 15C11
=
=
= 15 × 7 × 13
= 1365
Therefore, the total number of ways of choosing 11 players out of 15 is 1365.
Вопрос 2. Сколько различных партий лодок по 8 человек, состоящих из 5 мальчиков и 3 девочек, можно составить из 25 мальчиков и 10 девочек?
Решение:
We are given,
Total number of boys = 25
Total number of girls = 10
Boat party of 8 is to be made from 25 boys and 10 girls, by selecting 5 boys and 3 girls.
So, number of ways = 25C5 × 10C3
=
= 53130 × 120
= 6375600
Therefore, the total number of different boat parties that can be made is 6375600.
Вопрос 3. Сколькими способами студент может выбрать 5 курсов из 9 курсов, если 2 курса являются обязательными для каждого студента?
Решение:
We are given,
Total number of courses = 9
Number of courses a students can have = 5
Out of 9 courses, 2 courses are compulsory. So, a student can choose from 3 (i.e., 5−2) courses only.
So, number of ways = 7C3
=
=
= 7 × 5
= 35
Therefore, the total number of ways in which a student can choose the subjects is 35.
Вопрос 4. Сколькими способами можно выбрать футбольную команду из 11 игроков из 16? Сколько из них будет
(i) включить 2 конкретных игроков?
(ii) исключить 2 конкретных игроков?
Решение:
Total number of players = 16
Number of players to be selected = 11
So, number of ways = 16C11
=
=
= 4×7×13×12
= 4368
(i) include 2 particular players?
As two particular players have to be kept in the team every time, we have to choose 9 players (11−2) out of the
14 players (16−2).
So, number of ways = 14C9
=
=
= 7×13×11×2
= 2002
(ii) exclude 2 particular players?
As 2 particular players are already removed, we have to select 11 players out of the remaining 14 players (16−2).
So, number of ways = 14C9
=
=
= 14×13×2
= 364
Therefore, the required number of ways are 4368, 2002, 364 respectively.
Вопрос 5. Есть 10 профессоров и 20 студентов, из которых должна быть сформирована комиссия из 2 профессоров и 3 студентов. Найдите количество способов, которыми это можно сделать. Далее, найдите в скольких из этих комитетов:
(i) включен конкретный профессор.
(ii) конкретный студент включен.
(iii) конкретный учащийся исключается.
Решение:
We are given,
Total number of professor = 10
Total number of students = 20
It is given that a committee has to be formed by choosing 2 professors from 10 and 3 students from 20.
So, number of ways = 10C2 × 20C3
=
=
= 5×9×10×19×6
= 51300 ways
(i) a particular professor is included.
As one particular professor has to be selected in the committee every time, we have to choose 1 professor (2−1) out of the 9 professors (10−1). Number of ways for choosing students remain the same.
So, number of ways = 9C1 × 20C3
=
=
= 9×10×19×6
= 10260 ways
(ii) a particular student is included.
As one particular student has to be selected in the committee every time, we have to choose 2 professors (3−1) out of the 19 professors (20−1). Number of ways for choosing professors remain the same.
So, number of ways = 10C2 × 19C2
=
=
= 5×9×19×9
= 7695 ways
(iii) a particular student is excluded.
As one particular student has been removed from selection panel, we have to choose 3 students out of 19 students (20−1). Number of ways for choosing professors remain the same.
So, number of ways = 10C2 × 19C3
=
=
= 5×9×19×3×17
= 43605 ways
Therefore, the required number of ways are 51300, 10260, 7695, 43605 respectively.
Вопрос 6. Сколько различных произведений можно получить, умножив два или более чисел 3, 5, 7, 11 (без повторения)?
Решение:
Total number of ways will be sum of number of ways of multiplying two numbers, three numbers and four numbers.
So, number of ways = 4C2 + 4C3 + 4C4
=
=
= 6 + 4 + 1
= 11
Therefore, the total number of ways of product is 11 ways.
Вопрос 7. Из класса, состоящего из 12 мальчиков и 10 девочек, для конкурса выбрать 10 учеников, в том числе не менее 4 мальчиков и 4 девочек. 2 девушки, которые выиграли призы в прошлом году, должны быть включены. Сколькими способами можно сделать выбор?
Решение:
We are given,
Total number of boys = 12
Total number of girls = 10
Total number of girls for the competition = 10 + 2 = 12
As two particular girls have to be included, total students that can be selected = 10−2 = 8
So, number of ways = (12C6 × 8C2) + (12C5 × 8C3) + (12C4 × 8C4)
=
=
= (924 × 28) + (792 × 56) + (495 × 70)
= 25872 + 44352 + 34650
= 104874
Therefore, the total number of ways in which the selection can be made is 104874.
Вопрос 8. Сколько различных вариантов выбора 4 книг можно сделать из 10 разных книг, если
(i) нет ограничений
(ii) всегда выбираются две конкретные книги
(iii) никогда не выбираются две конкретные книги
Решение:
Общее количество книг = 10
Количество книг для выбора = 4
(i) нет ограничений
Number of ways = Choosing 4 books out of 10 books = 10C4
=
=
= 10×3×7
= 210
Therefore, the number of ways of selecting books if there is no restriction is 210.
(ii) всегда выбираются две конкретные книги
As we have to select two particular books every time, we can select 2 books (4−2) out of the remaining 8 books (10−2).
So, number of ways = 8C2
=
=
= 4×7
= 28
Therefore, the number of ways of selecting books if two particular books are always selected is 28.
(iii) никогда не выбираются две конкретные книги
As two particular books have been removed, we have to choose 4 books out of the remaining 8 books (10−2).
So, number of ways = 8C4
=
=
= 7×2×5
= 70
Therefore, the number of ways of selecting books if two particular books are never selected is 70.
Вопрос 9. Сколькими способами из 4 офицеров и 8 джаванов можно выбрать 6
(i) включить ровно одного офицера
(ii) включить хотя бы одного офицера?
Решение:
Общее количество офицеров = 4
Общее количество джаванов = 8
Общее количество выборов, которые нужно сделать = 6
(i) включить ровно одного офицера
Out of 6 selections, only 1 has to be an officer. So remaining 5 have to be jawans.
So, number of ways = (4C1) × (8C5)
=
=
= 4×4×7×2
= 224
Therefore, the number of ways of selection if only one officer has to be included is 224.
(ii) включить хотя бы одного офицера?
Out of 6 selections, at least 1 has to be an officer. So, we can choose from 1 to all 4 officers in our selections. And number of jawans would adjust according to that.
So, number of ways = (4C1 × 8C5) + (4C2 × 8C4) + (4C3 × 8C3) + (4C4 × 8C2)
=
=
= (4 × 56) + (6 × 70) + (4 × 56) + (1 × 28)
= 224 + 420 + 224 + 28
= 896
Therefore, the number of ways of selection if at least one officer has to be included is 896.
Вопрос 10. Сформировать спортивную команду из 11 учащихся, выбрав не менее 5 человек из XI класса и не менее 5 человек из XII класса. Если в каждом из этих классов по 20 учеников, сколькими способами можно составить команды?
Решение:
Total number of students in XI = 20
Total number of students in XII = 20
Number of students to be selected in a team = 11
Now, at least 5 from class XI and 5 from class XII have to be selected.
So, number of ways = (20C6 × 20C5) + (20C5 × 20C6)
= 2 (20C6 × 20C5)
= 2
=
= 2×38760×15504
= 1201870080
Therefore, the number of ways in which the teams can be constituted is 1201870080.
Вопрос 11. Студент должен ответить на 10 вопросов, выбрав не менее 4 из каждой части А и части Б. Если в части А 6 вопросов, а в части Б 7, сколькими способами студент может выбрать 10 вопросов?
Решение:
Total number of questions = 10
Questions in part A = 6
Questions in part B = 7
Number of questions a student can choose = 10
Now a student can choose at least 4 from each of part A and part B and total number of questions that he can choose must not exceed.
So, number of ways = (6C4 × 7C6) + (6C5 × 7C5) + (6C6 × 7C4)
=
=
= (15×7) + (6×21) + (1×35)
= 105 + 126 + 35
= 266
Therefore, the number of ways of answering 10 questions is 266.
