第11类RD Sharma解决方案–第17章组合-练习17.2 |套装1

问题1.从15名板球运动员中选出11名运动员。有多少种方法可以做到这一点?

解决方案：

We are given,

Total number of players = 15

Number of players to be chosen = 11

So, number of ways = ¹⁵C₁₁

=$\frac{15!}{11!4!}$

=$\frac{15×14×13×12}{4×3×2×1}$

= 15 × 7 × 13

= 1365

Therefore, the total number of ways of choosing 11 players out of 15 is 1365.

问题2：25个男孩和10个女孩可以组成8个不同的乘船聚会，包括5个男孩和3个女孩?

解决方案：

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

=$\frac{25!}{5!20!}×\frac{10!}{3!7!}$

= 53130 × 120

= 6375600

Therefore, the total number of different boat parties that can be made is 6375600.

问题3.如果每个学生都必须修读2门课程，那么学生可以从9种课程中选择5门课程?

解决方案：

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

=$\frac{7!}{3!4!}$

=$\frac{7×6×5}{3×2×1}$

= 7 × 5

= 35

Therefore, the total number of ways in which a student can choose the subjects is 35.

问题4.从16名球员中选出11名球员的足球队有多少种方式?其中有多少将

(i)包括2名特定球员?

(ii)排除2位特定玩家?

解决方案：

Total number of players = 16

Number of players to be selected = 11

So, number of ways = ¹⁶C₁₁

=$\frac{16!}{11!5!}$

=$\frac{16×15×14×13×12}{5×4×3×2×1}$

= 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

=$\frac{14!}{9!5!}$

=$\frac{14×13×12×11×10}{5×4×3×2×1}$

= 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

=$\frac{14!}{11!3!}$

=$\frac{14×13×12}{3×2×1}$

= 14×13×2

= 364

Therefore, the required number of ways are 4368, 2002, 364 respectively.

问题5：有10名教授和20名学生，其中将组成一个由2名教授和3名学生组成的委员会。查找完成此操作的方法数量。进一步，找出这些委员会中有多少个：

(i)包括一名特定的教授。

(ii)包括一个特定的学生。

(iii)排除特定学生。

解决方案：

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

=$\frac{10!}{2!8!}×\frac{20!}{3!17!}$

=$\frac{10×9}{2×1}×\frac{20×19×18}{3×2×1}$

= 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

=$\frac{9!}{8!}×\frac{20!}{3!17!}$

=$\frac{9×8!}{8!}×\frac{20×19×18}{3×2×1}$

= 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

=$\frac{10!}{2!8!}×\frac{19!}{2!17!}$

=$\frac{10×9}{2×1}×\frac{19×18}{2×1}$

= 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

=$\frac{10!}{2!8!}×\frac{19!}{3!16!}$

=$\frac{10×9}{2×1}×\frac{19×18×17}{3×2×1}$

= 5×9×19×3×17

= 43605 ways

Therefore, the required number of ways are 51300, 10260, 7695, 43605 respectively.

问题6.将3、5、7、11中的两个或多个数字相乘(无重复)可以得到多少个不同的乘积?

解决方案：

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

=$\frac{4!}{2!2!}+\frac{4!}{3!}+\frac{4!}{4!}$

=$\frac{4×3}{2×1}+\frac{4}{1}+1$

= 6 + 4 + 1

= 11

Therefore, the total number of ways of product is 11 ways.

问题7.从12名男孩和10名女孩的班级中，将选择10名学生参加比赛，至少包括4名男孩和4名女孩。去年获奖的两个女孩应该包括在内。可以通过几种方式进行选择?

解决方案：

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)

=$(\frac{12!}{6!6!}×\frac{8!}{2!6!})+(\frac{12!}{7!5!}×\frac{8!}{5!3!})+(\frac{12!}{8!4!}×\frac{8!}{4!4!})$

=$(\frac{12×11×10×9×8×7}{6×5×4×3×2×1}×\frac{8×7}{2×1})+(\frac{12×11×10×9×8}{5×4×3×2×1}×\frac{8×7×6}{3×2×1})+(\frac{12×11×10×9}{4×3×2×1}×\frac{8×7×6×5}{4×3×2×1})$

= (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.

问题8.如果有10种不同的书籍，可以从4种书籍中选择多少种?

(i)没有限制

(ii)总是选择两本特定的书

(iii)从未选择过两本特别的书

解决方案：

总书数= 10

所选书籍数量= 4

(i)没有限制

Number of ways = Choosing 4 books out of 10 books = ¹⁰C₄

=$\frac{10!}{4!6!}$

=$\frac{10×9×8×7}{4×3×2×1}$

= 10×3×7

= 210

Therefore, the number of ways of selecting books if there is no restriction is 210.

(ii)总是选择两本特定的书

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

=$\frac{8!}{2!6!}$

=$\frac{8×7}{2×1}$

= 4×7

= 28

Therefore, the number of ways of selecting books if two particular books are always selected is 28.

(iii)从未选择过两本特别的书

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

=$\frac{8!}{4!4!}$

=$\frac{8×7×6×5}{4×3×2×1}$

= 7×2×5

= 70

Therefore, the number of ways of selecting books if two particular books are never selected is 70.

问题9：从4名军官和8名颚动物中，有多少种可以选择6种

(i)恰好包括一名人员

(ii)包括至少一名军官?

解决方案：

警员总数= 4

颚骨总数= 8

选择总数= 6

(i)恰好包括一名人员

Out of 6 selections, only 1 has to be an officer. So remaining 5 have to be jawans.

So, number of ways = (⁴C₁) × (⁸C₅)

=$\frac{4!}{3!1!}×\frac{8!}{3!5!}$

=$\frac{4}{1}×\frac{8×7×6}{3×2×1}$

= 4×4×7×2

= 224

Therefore, the number of ways of selection if only one officer has to be included is 224.

(ii)包括至少一名军官?

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)

=$(\frac{4!}{1!3!}×\frac{8!}{5!3!})+(\frac{4!}{2!2!}×\frac{8!}{4!4!})+(\frac{4!}{3!1!}×\frac{8!}{3!5!})+(\frac{4!}{4!0!}×\frac{8!}{2!6!})$

=$(\frac{4}{1}×\frac{8×7×6}{3×2×1})+(\frac{4×3}{2×1}×\frac{8×7×6×5}{4×3×2×1})+(\frac{4}{1}×\frac{8×7×6}{3×2×1})+(\frac{1}{1}×\frac{8×7}{2×1})$

= (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.

问题10：将组成一个由11名学生组成的运动队，从第XI课中至少选择5人，从第XII课中至少选择5人。如果每个班级有20名学生，可以用多少种方式组成小组?

解决方案：

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$(\frac{20!}{6!14!}×\frac{20!}{5!15!})$

=$2×\frac{20×19×18×17×16×15}{6×5×4×3×2×1}×\frac{20×19×18×17×16}{5×4×3×2×1}$

= 2×38760×15504

= 1201870080

Therefore, the number of ways in which the teams can be constituted is 1201870080.

问题11.学生必须回答10个问题，从A部分和B部分中至少选择4个。如果A部分有6个问题，B部分有7个问题，则学生可以选择多少种方式?

解决方案：

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)

=$(\frac{6!}{4!2!}×\frac{7!}{6!1!})+(\frac{6!}{5!1!}×\frac{7!}{5!2!})+(\frac{6!}{6!0!}×\frac{7!}{4!3!})$

=$(\frac{6×5}{2×1}×\frac{7}{1})+(\frac{6}{1}×\frac{7×6}{2×1})+(\frac{1}{1}×\frac{7×6×5}{3×2×1})$

= (15×7) + (6×21) + (1×35)

= 105 + 126 + 35

= 266

Therefore, the number of ways of answering 10 questions is 266.