NCERT Solutions for Exercise 7.4 Class 11 Maths Chapter 7 - Permutations and Combinations

# NCERT Solutions for Exercise 7.4 Class 11 Maths Chapter 7 - Permutations and Combinations

Edited By Vishal kumar | Updated on Nov 07, 2023 01:27 PM IST

## NCERT Solutions for Class 11 Maths Chapter 7 Permutations And Combinations Exercise 7.4- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 7: Permutations and Combinations Exercise 7.4- In the previous exercises of this chapter, you have already learned about the permutations when objects are distinct, permutations when the objects are not distinct, etc. In the NCERT solutions for Class 11 Maths Chapter 7 Exercise 7.4, you will learn about the combinations. If you have a command on permutations it won't take much effort to get a command on the Class 11 Maths chapter 7 exercise 7.4.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

Scholarship Test: Vidyamandir Intellect Quest (VIQ)

Both topics permutations and combinations are very important concepts useful in other topics such as probability. These concepts are very useful to solve some real-life problems like selecting fruits out of some given fruits, and selecting n best books out of N books for the same subject. This is the new topic added in Class 11 which hasn't been studied in the previous classes, so you may find it difficult to understand. You should try to solve more problems in order to get conceptual clarity. If you are not able to solve NCERT problems at first by yourself, you can go through the Class 11 Maths chapter 7 exercise 7.4 solutions. Check the NCERT solutions link to get NCERT solutions for C lasses 6 to 12 at one place.

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**As per the CBSE Syllabus for 2023-24, this chapter has been renumbered as Chapter 6.

## Question:1 If $^{n}C_{8}=^{n}\; \! \! \! \! C_{2},$ find $^{n}C_{2}$

Given : $^{n}C_{8}=^{n}\; \! \! \! \! C_{2},$

We know that $^nC_a=^nC_b$ $\Rightarrow a=b\, \, \, or\, \, n=a+b$

$^{n}C_{8}=^{n}\; \! \! \! \! C_{2},$

$\Rightarrow n=8+2$

$\Rightarrow n=10$

$^{n}C_{2}$$=^{10}C_2$

$=\frac{10!}{(10-2)!2!}$

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

$=\frac{10\times 9\times 8!}{8!2!}$

$=5\times 9=45$

Question:2(i) Determine n if

$\; ^{2n}C_{3}:^{n}C_{3}=12:1$

Given that : $(i)\; ^{2n}C_{3}:^{n}C_{3}=12:1$

$\Rightarrow \, \, \frac{^{2n}C_3}{^nC_3}=\frac{12}{1}$

The ratio can be written as

$\Rightarrow \, \, \frac{\frac{2n!}{(2n-3)!3!}}{\frac{n!}{(n-3)!3!}}=\frac{12}{1}$

$\Rightarrow \, \, {\frac{2n!(n-3)!}{(2n-3)!n!}}=\frac{12}{1}$

$\Rightarrow \, \, \frac{2n\times (2n-1)\times (2n-2)\times (2n-3)!\times (n-3)!}{(2n-3)!\times n\times (n-1)\times(n-2)\times (n-3)! }=\frac{12}{1}$

$\Rightarrow \, \, \frac{2n\times (2n-1)\times (2n-2)}{ n\times (n-1)\times(n-2) }=\frac{12}{1}$

$\Rightarrow \, \, \frac{2\times (2n-1)\times 2}{ (n-2) }=\frac{12}{1}$

$\Rightarrow \, \, \frac{4\times (2n-1)}{ (n-2) }=\frac{12}{1}$

$\Rightarrow \, \, \frac{ (2n-1)}{ (n-2) }=\frac{3}{1}$

$\Rightarrow \, \, 2n-1=3n-6$

$\Rightarrow \, \, 6-1=3n-2n$

$\Rightarrow \, \, n=5$

Question:2(ii) Determine n if

$\; ^{2n}C_{3}:^{n}C_{3}=11:1$

Given that : $(ii)\; ^{2n}C_{3}:^{n}C_{3}=11:1$

$\Rightarrow \, \, \frac{^{2n}C_3}{^nC_3}=\frac{11}{1}$

$\Rightarrow \, \, \frac{\frac{2n!}{(2n-3)!3!}}{\frac{n!}{(n-3)!3!}}=\frac{11}{1}$

$\Rightarrow \, \, {\frac{2n!(n-3)!}{(2n-3)!n!}}=\frac{11}{1}$

$\Rightarrow \, \, \frac{2n\times (2n-1)\times (2n-2)\times (2n-3)!\times (n-3)!}{(2n-3)!\times n\times (n-1)\times(n-2)\times (n-3)! }=\frac{11}{1}$

$\Rightarrow \, \, \frac{2n\times (2n-1)\times (2n-2)}{ n\times (n-1)\times(n-2) }=\frac{11}{1}$

$\Rightarrow \, \, \frac{2\times (2n-1)\times 2}{ (n-2) }=\frac{11}{1}$

$\Rightarrow \, \, \frac{4\times (2n-1)}{ (n-2) }=\frac{11}{1}$

$\Rightarrow \, \, 8n-4=11n-22$

$\Rightarrow \, \, 22-4=11n-8n$

$\Rightarrow \, \, 3n=18$

$\Rightarrow \, \, n=6$

Thus the value of n=6

To draw chords 2 points are required on the circle.

To know the number of chords on the circle , when points on the circle are 21.

Combinations =Number of chords $=^{21}C_2$

$=\frac{21!}{(21-2)!2!}$

$=\frac{21!}{19!2!}$

$=\frac{21\times 20}{2}$

$=210$

A team of 3 boys and 3 girls be selected from 5 boys and 4 girls.

3 boys can be selected from 5 boys in $^5C_3$ ways.

3 girls can be selected from 4 boys in $^4C_3$ ways.

Therefore, by the multiplication principle, the number of ways in which a team of 3 boys and 3 girls can be selected $=^5C_3\times ^4C_3$

$=\frac{5!}{2!3!}\times \frac{4!}{1!3!}$

$=10\times 4=40$

There are 6 red balls, 5 white balls and 5 blue balls.

9 balls have to be selected in such a way that consists of 3 balls of each colour.

3 balls are selected from 6 red balls in $^6C_3$.

3 balls are selected from 5 white balls in $^5C_3$

3 balls are selected from 5 blue balls in $^5C_3$.

Hence, by the multiplication principle, the number of ways of selecting 9 balls $=^6C_3\times ^5C_3\times ^5C_3$

$=\frac{6!}{3!3!}\times \frac{5!}{2!3!}\times \frac{5!}{2!3!}$

$=20\times 10\times 10=2000$

In a deck, there is 4 ace out of 52 cards.

A combination of 5 cards is to be selected containing exactly one ace.

Then, one ace can be selected in $^4C_1$ ways and other 4 cards can be selected in $^4^8C_4$ ways.

Hence, using the multiplication principle, required the number of 5 card combination $=^4C_1\times ^4^8C_4$

$=\frac{4!}{1!3!}\times \frac{48!}{44!4!}$

$=4\times \frac{48\times 47\times 46\times 45}{4\times 3\times 2}=778320$

Out off, 17 players, 5 are bowlers.

A cricket team of 11 is to be selected such that there are exactly 4 bowlers.

4 bowlers can be selected in $^5C_4$ ways and 7 players can be selected in $^1^2C_7$ ways.

Thus, using multiplication priciple, number of ways of selecting the team $=^5C_4 .^1^2C_7$

$=\frac{5!}{1!4!}\times \frac{12!}{5!7!}$

$=5 \times \frac{12\times 11\times 10\times 9\times 8}{5\times 4\times 3\times 2}$

$=3960$

A bag contains 5 black and 6 red balls.

2 black balls can be selected in $^5C_2$ ways and 3 red balls can be selected in $^6C_3$ ways.

Thus, using multiplication priciple, number of ways of selecting 2 black and 3 red balls $=^5C_2 .^6C_3$

$=\frac{5!}{2!3!}\times \frac{6!}{3!3!}$

$=10\times 20=200$

9 courses are available and 2 specific courses are compulsory for every student.

Therefore, every student has to select 3 courses out of the remaining 7 courses.

This can be selected in $^7C_3$ ways.

Thus, using multiplication priciple, number of ways of selecting courses $=^7C_3$

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

$= \frac{7\times 6\times 5}{ 3\times 2}$

$=35$

## More About NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.4:-

Class 11 Maths chapter 7 exercise 7.4 consists of questions related to finding the number of ways to select n number of things out of N things. This selection of objects is called a combination in the bookish language. There are three examples and a couple of theorems related to combinations are given in the NCERT textbook before the Class 11 Maths chapter 7 exercise 7.4. You must go through the proof of these theorems in order to get conceptual clarity. There are nine questions given in the Class 11 Maths chapter 7 exercise 7.4 that you must solve by yourself.

## Benefits of NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.14:-

• Exercise 7.4 Class 11 Maths is useful for solving real-life problems also.
• Class 11 Maths chapter 7 exercise 7.4 solutions are designed by subject matter experts, so you can rely upon these solutions.
• You must have conceptual clarity about the Class 11th Maths chapter 7 exercise 7.4 to understand probability.
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## Key Features of NCERT 11th Class Maths Exercise 7.4 Answers

1. In-Depth Coverage: The 11th class maths exercise 7.4 answers comprehensively address all the exercises problems found in exeercise 7.4 of the NCERT 11th Class Mathematics textbook.

2. Detailed Explanations: The ex 7.4 class 11 solutions are presented with meticulous explanations, ensuring that students can understand the problem-solving process with clarity.

3. Precision and Simplicity: The class 11 maths ex 7.4 solution are crafted with precision and written in a simple and straightforward language to make the concepts easily understandable for students.

4. Correct Mathematical Notation: These class 11 ex 7.4 solutions employ the appropriate mathematical notations and terminology, enabling students to become fluent in mathematical language.

5. Fostering Conceptual Understanding: The 11th class maths exercise 7.4 answers solutions are designed to deepen students' grasp of mathematical concepts, promoting critical thinking and enhancing their problem-solving skills.

6. No Cost Access: Typically, these ex 7.4 class 11 answers are accessible at no charge, providing students with the freedom to access and use them whenever needed.

7. Supplementary Learning Resource: These class 11 ex 7.4 answers can serve as valuable supplementary materials to reinforce classroom teachings and support students in their exam preparation.

8. Homework and Practice: Students can utilize these answers to cross-check their work, practice solving problems, and improve their overall performance in mathematics.

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## Subject Wise NCERT Exampler Solutions

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1. What is the weightage of the probability in JEE Main Maths ?

Generally one question is asked from probability in the Maths portion of JEE Main exam which means it has 3.3% weightage in the JEE Main Maths.

2. Where can I get syllabus for Class 11 Maths ?

Here you will get Syllabus for Class 11 Maths.

3. Can I get NCERT solutions for Class 11 Maths for free ?

4. Where can I get NCERT exemplar solutions for Class 11 Maths ?
5. What is the weightage of the mathematical reasoning in the CBSE Class 11 Maths ?

Mathematical reasoning has 2 marks weightage in the CBSE Class 11 Maths.

6. What is the weighatge of limits in JEE Main Maths ?

Generally one question is asked from limit in the JEE Main Maths which means it has 3.3% weightage in the JEE Main Maths.

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