第11类RD Sharma解决方案–第17章组合-练习17.2 |套装2

问题12.在考试中，学生要回答5个问题中的4个问题；但是，问题1和2是强制性的。确定学生做出选择的方式数量。

解决方案：

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

=$\frac{3!}{2!1!}$

=$\frac{3×2×1}{2×1}$

= 3

Therefore, the number of ways answering the questions is 3.

问题13.候选人必须回答12个问题中的7个问题，该问题分为两组，每组包含6个问题。不允许他在任何一组中尝试超过5个问题。他可以从几种方式中选择7个问题?

解决方案：

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)

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

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

= (6×15) + (15×20) + (20×15) + (15×6)

= 90 + 300 + 300 + 90

= 780

Therefore, the number of ways of answering 7 questions is 780.

问题14.平面中有10个点，其中4个点是共线的。通过连接这些点可以绘制多少条不同的直线。

解决方案：

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

=$\frac{10!}{2!8!}-\frac{4!}{2!2!}+1$

=$\frac{10×9}{2×1}-\frac{4×3}{2×1}+1$

= 45 – 6 + 1

= 40

Therefore, the total number of ways of drawing different lines is 40.

问题15：求出对角线的数量

(i)六角形

解决方案：

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

=$\frac{6!}{2!4!}$

=$\frac{6×5}{2×1}$

= 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)16边的多边形

解决方案：

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

=$\frac{16!}{2!14!}$

=$\frac{16×15}{2×1}$

= 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.

问题16.通过连接12个点可以获得多少个三角形，其中五个是共线的?

解决方案：

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

=$\frac{12!}{3!9!}-\frac{5!}{3!2!}$

=$\frac{12×11×10}{3×2×1}-\frac{5×4}{2×1}$

= (2×11×10) – (5×2)

= 220 – 10

= 210

Therefore, the total number of triangles that can be formed is 210.

问题17：在必须选出至少一名妇女的情况下，可以由多少种方式组成一个由6名男子和4名妇女组成的5人委员会?

解决方案：

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)

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

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

= 60 + 120 + 60 + 6

= 246

Therefore, the number of ways of selection is 246.

问题18：在一个村庄里，有87个家庭，其中52个家庭最多有两个孩子。在一项农村发展计划中，将选择20个家庭作为援助对象，其中18个家庭最多必须有2个孩子。可以通过几种方式进行选择?

解决方案：

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)

问题19：一组由4名女孩和7名男孩组成。如果一个团队有5个成员，则可以通过几种方式选择该团队

(i)没有女孩

(ii)至少一个男孩和女孩

(iii)至少3个女孩

解决方案：

女孩数= 4

男孩人数= 7

要选择的成员数= 5

(i)没有女孩

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

=$\frac{7!}{5!2!}$

=$\frac{7×6}{2×1}$

= 21

Therefore, the number of ways of selection such that team has no girls is 21.

(ii)至少一个男孩和女孩

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)

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

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

= 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)至少3个女孩

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)

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

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

= 84 + 7

= 91

Therefore, the number of ways of selection such that team has at least 3 girls is 91.

问题20：一个由3名男子和2名妇女组成的小组，由3人组成。有多少种方法可以做到这一点?这些委员会中有多少个由1名男性和2名女性组成?

解决方案：

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

=$\frac{5!}{3!2!}$

=$\frac{5×4}{2×1}$

= 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

=$\frac{2!}{1!1!}×\frac{3!}{2!1!}$

= 6

问题21.找出

(i)在十边形中形成的对角线。

解决方案：

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

=$\frac{10!}{2!8!}$

=$\frac{10×9}{2×1}$

= 45

Number of sides = 10

So, number of diagonals = 45−10 = 35.

Therefore, number of diagonals formed in a decagon is 35.

(ii)在十边形中形成的三角形。

解决方案：

A decagon has 10 sides. By joining any 3 angular points we get a triangle.

So number of lines formed = ¹⁰C₃

=$\frac{10!}{3!7!}$

=$\frac{10×9×8}{3×2×1}$

=$\frac{720}{6}$

= 120

Therefore, number of triangles formed in a decagon is 120.

问题22.如果5张卡中的至少一张必须是国王，请确定52张卡中的5张卡组合的数量?

解决方案：

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)

=$(\frac{4!}{1!3!}×\frac{48!}{4!44!})+(\frac{4!}{2!2!}×\frac{48!}{3!45!})+(\frac{4!}{3!1!}×\frac{48!}{2!46!})+(\frac{4!}{4!0!}×\frac{48!}{1!47!})$

=$(\frac{4}{1}×\frac{48×47×46×45}{4×3×2×1})+(\frac{4×3}{2×1}×\frac{48×47×46}{3×2×1})+(\frac{4}{1}×\frac{48×47}{2×1})+(\frac{1}{1}×\frac{48}{1})$

= 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.