# NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations (Ex 7.4) Exercise 7.4

Students of Class 11 are introduced to some new concepts after they pass out of Class 10 and choose to study mathematics in Class 11. Students of Class 11 are first challenged with choosing a path for themselves by choosing the subjects of their interest. This could be difficult for some students, as not every student has clarity after Class 10 about what they want to study. But students can find their interest in things after exploring their interests themselves. They can choose any subject they want to study for Class 11 and 12. They can also study these subjects further at universities.

Students must remember that studying with focus and concentration will help them gain in-depth knowledge of the chapter. This will help them build a foundation of knowledge about the subject and the concept. Having a strong knowledge of the basics of a subject and topic will help them solve advanced questions about that subject and topic. The NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4, will be helpful to students in this regard. The NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4 will help students to form a basic understanding of this concept.

## NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations (Ex 7.4) Exercise 7.4

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### Exercise 7.4 of Chapter 7- Permutations and Combinations is based on the following topic

The NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4 provides an overview of the Permutation and Combination Exercise 7.4. The Class 11 Maths Chapter 7 Exercise 7.4 is based on the Combinations part of the chapter. This exercise covers the topics about combinations like how it is different from permutation and how it is solved. This exercise also covers topics such as theorems based on Combinations and the properties of combinations. Some formulas for combinations have also been derived before Exercise 7.4 Class 11 to help students understand it better. Deriving a formula can help students to understand the difference between combinations and permutations and in which cases the formula should be applied. Though, the NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4 can also help students with that.

The NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4, will help students get a better understanding of Combinations. They must remember that Combinations are an important part of their syllabus and this chapter. Studying combinations with focus is important for students, as that will help them solve further mathematical problems. It will also help them in solving miscellaneous exercises in this chapter. The miscellaneous exercise contains questions from all the topics studied by students in this chapter. Students could find it challenging to figure out which formula has to be applied for which chapter. It could be challenging for students to figure out which questions require concepts and formulas from Permutation and which ones need the formula from Combination. Hence, students must use the NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4 to understand the topic better and faster.

Studying and practising NCERT could help students get past these problems. It is important to do the exercises from the start instead of jumping to the miscellaneous exercises first. This could help students make fewer mistakes and understand concepts better before they start applying these formulas. Students can get NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4, on the Extramarks website and app to help with their preparation. It is important to study every concept of a chapter as it might not seem important at the time but it could have an application in other questions of different exercises and chapters.

### Access NCERT Solutions for Maths Class 11 Chapter 7 – Permutations and Combinations

The NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4, have been made available to students on the Extramarks website and app. Students can access the NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4, whenever they want while they are studying. Students need to understand the importance of practising Mathematics. Students might be making unintentional errors in their answers. These mistakes can end up making students get wrong answers to the question even though they have used the correct method of solution. Making such mistakes can be avoided by students if they practice enough. They must understand that making mistakes is a natural part of learning, but they must identify and correct their mistakes to get better at it. The NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4 can help students with this.

To avoid making mistakes, practising is required as it will help students identify their errors and be conscious of them during the exam. The correct solutions can be found on the NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4. Students must remember that the NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4 is provided so students can practice what they have learnt during their classes and during the time they studied these concepts.

### NCERT Solutions for Class 11 Maths Chapters

NCERT solutions to Exercise 7.4 Class11 are available on the Extramarks website and app. NCERT Solutions are important for students to study from as they will help them understand the topics. Students must remember that solving NCERT books is important because it can help students score better in both their board exams and entrance exams like JEE, CUET, etc., as NCERT can help strengthen their basic understanding of the chapter and also help them understand how the questions must be answered. They can get help related to that from Extramarks’ NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4.

Therefore, students must practice all the chapters from NCERT, as it is an important part of the preparation for exams. If they need help with anything, they can look at the NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4. They can find academic help in the NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4, for any question they have a doubt about. Students can get their doubts solved by the subject matter experts at Extramarks.

Students must look at all the example questions and the exercise questions given in the NCERT to have in-depth knowledge of the chapter. This is an important part of the preparation for any exam, whether the end-of-term exam or the entrance exam. If students find it hard to understand the methods of solutions provided to them by NCERT through the example questions, they can refer to the NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4 by Extramarks for studying.

### NCERT Solution Class 11 Maths of Chapter 7 Exercise

The Chapter Permutations and Combinations includes topics such as the fundamental principle of counting, permutations, permutations when all the objects are distinct, permutations when all the objects are not distinct, and combinations. These topics are new for students, so they might not be able to start solving these questions instantly, but with good practice, they can understand and solve the questions. Students must remember that they can take help from the NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4.

The Chapter – Pemutations and Combinations requires students to understand the concept before they start solving the questions. Students will not be able to solve the question with the basic knowledge they already have. They need to have in-depth knowledge of the chapter to be able to solve advanced problems for the entrance exam student may be preparing for.

The provided NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4, are supposed to help students find easier and faster ways of solving the questions. This way, students will be able to solve the questions and save time by checking for any mistakes they have made in any question and correct them. The NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4 will help students figure out how to solve the question without making any mistakes. The questions will help students later solve more advanced problems. This will reflect positively on their academic performance. Good academics will help students get into the university of their choice and the programme they are interested in. The NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4, will be of help to students in that regard.

Exercise 7.4 is followed by the miscellaneous exercise, which includes questions from all the topics studied so far in the chapter. Students could have difficulty solving it if they have not studied previous exercises thoroughly. Hence, students must focus on what they are studying to get a good understanding of the chapter and be able to solve advanced questions. The NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4 will help students with this reference.

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

The NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4 will help students overcome their doubts about the chapter and mistakes in the answers. The students must prepare for NCERT to get a good score in their exams. It is important for their academic performance. Students must remember what they have learnt before by having enough time for revision. The NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4 could help students save time for revision. The NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4 could also help students at the time of the revision, as it provides an overview of the chapter. It can help students remember the method of solution for specific questions, as some questions could be tricky and students might get stuck while solving them if they do not remember the proper method used to solve them. The NCERT Solutions For Class 11 Maths Chapter 7 Exercise 7.4 can help students with that problem.

Students must prepare for the exam with full concentration so that they do not face any difficulty during the exam. They need to work on improving their quality of study so they make fewer mistakes during examinations. Being aware of how to write an examination can help them keep up the good work. Taking breaks during their study time will help them stay healthy and motivated for their studies. This way, they will get time for extracurricular activities and hobbies. Doing things apart from studying will affect their mood positively. This way, they will study better and with more focus. This all could result in them improving their academic performance and scores and help them develop personally too.

**Q.1 ** If ^{n}C_{8} = ^{n}C_{2}, find ^{n}C_{2}.

**Ans.**

$\mathrm{Given}{,}^{\text{n}}{\text{C}}_{\text{8}}={\text{\hspace{0.17em}}}^{\text{n}}{\text{C}}_{\text{2}}$

$\begin{array}{l}{\Rightarrow}^{\text{n}}{\text{C}}_{\text{8}}={\text{\hspace{0.17em}}}^{\text{n}}{\text{C}}_{\text{n-2}}\text{}[\because {\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-\mathrm{r}}]\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8=\mathrm{n}-2\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}10=\mathrm{n}\\ \mathrm{So},{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}={\text{\hspace{0.17em}}}^{10}{\mathrm{C}}_{2}\\ =\frac{10!}{2!(10-2)!}\\ =\frac{10\times 9\times 8!}{2!\text{\hspace{0.17em}}8!}\\ =45\end{array}$

**Q.2 ** Determine n if

(i) ^{2n}C_{3 }: ^{n}C_{3} = 12 : 1

(ii) ^{2n}C_{3} : ^{n}C_{3} = 11 : 1

**Ans.**

$\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.33em}}\frac{{}^{\text{2n}}{\text{C}}_{\text{3}}}{{}^{\text{n}}{\text{C}}_{\text{3}}}=\frac{12}{1}\\ \Rightarrow \text{\hspace{0.33em}}\frac{\frac{2\text{n}!}{3!\left(2\text{n}-3\right)!}}{\frac{\text{n}!}{3!\left(\text{n}-3\right)!}}=12\\ \Rightarrow \text{\hspace{0.33em}}\frac{2\text{n}!}{3!\left(2\text{n}-3\right)!}\times \frac{3!\left(\text{n}-3\right)!}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)\left(\text{n}-3\right)!}=12\\ \Rightarrow \text{\hspace{0.33em}}\frac{2\text{n}\left(2\text{n}-1\right)\left(2\text{n}-2\right)\left(2\text{n}-3\right)!}{\left(2\text{n}-3\right)!}\times \frac{1}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)}=12\\ \Rightarrow \text{\hspace{0.33em}}2\text{n}\left(2\text{n}-1\right)2\left(\text{n}-1\right)\times \frac{1}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)}=12\\ \Rightarrow \text{\hspace{0.33em}}\frac{\left(2\text{n}-1\right)}{\text{n}-2}=3\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}2\text{n}-1=3\text{n}-6\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}6-1=3\text{n}-2\text{n}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}5=\text{n}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}n}=5.\\ \left(\text{ii}\right){\text{\hspace{0.33em}\hspace{0.33em}}}^{\text{2n}}{\text{C}}_{\text{3}}:{\text{\hspace{0.33em}}}^{\text{n}}{\text{C}}_{\text{3}}=\text{11}:\text{1}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{{}^{\text{2n}}{\text{C}}_{\text{3}}}{{}^{\text{n}}{\text{C}}_{\text{3}}}=\frac{11}{1}\\ \Rightarrow \text{\hspace{0.33em}}\frac{\frac{2\text{n}!}{3!\left(2\text{n}-3\right)!}}{\frac{\text{n}!}{3!\left(\text{n}-3\right)!}}=11\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{2\text{n}!}{3!\left(2\text{n}-3\right)!}\times \frac{3!\left(\text{n}-3\right)!}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)\left(\text{n}-3\right)!}=11\\ \Rightarrow \text{\hspace{0.33em}}\frac{2\text{n}\left(2\text{n}-1\right)\left(2\text{n}-2\right)\left(2\text{n}-3\right)!}{\left(2\text{n}-3\right)!}\times \frac{1}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)}=11\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}2\text{n}\left(2\text{n}-1\right)2\left(\text{n}-1\right)\times \frac{1}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)}=11\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{4\left(2\text{n}-1\right)}{\text{n}-2}=11\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}8\text{n}-4=11\text{n}-22\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}22-4=11\text{n}-8\text{n}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}18=3\text{n}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}n}=6\end{array}$

**Q.3 ** How many chords can be drawn through 21 points on a circle?

**Ans.**

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of points on a circle}=\text{21}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Number of points used for a chord}=\text{2}\\ \mathrm{Number}\text{of chords made by 21 points}={\text{\hspace{0.17em}}}^{21}{\mathrm{C}}_{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{21!}{2!(21-2)!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{21\times 20\times 19!}{2\times 1\times 19!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=210\\ \mathrm{Thus},\text{ number of chords in a circle is 210.}\end{array}$

**Q.4 ** In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

**Ans.**

$\begin{array}{l}\mathrm{Total}\text{ number of boys}=\text{5}\\ \text{Total number of girls}=4\\ \mathrm{Number}\text{ of boys in a team}=\text{3}\\ \text{Number of girls in a team}=\text{3}\end{array}$

$\begin{array}{l}\text{Number of ways to select 3 boys}={\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{3}\\ =\frac{5!}{3!(5-3)!}\\ =\frac{5\times 4\times 3!}{3!\text{\hspace{0.17em}}\times 2!}\\ =10\\ \text{Number of ways to select 3 girls}={\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{3}\end{array}$ \begin{array}{l}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}

$\begin{array}{l}=\frac{4!}{3!(4-3)!}\\ =\frac{4\times 3!}{3!\text{\hspace{0.17em}}\times 1!}\\ =4\\ \mathrm{Total}\text{\hspace{0.17em}number of ways to select 3}\\ \text{boys and 3 girls}=10\times 4\\ =40\end{array}$

**Q.5 ** Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

**Ans.**

$\begin{array}{l}\mathrm{Number}\text{of red balls}=\text{6}\\ \mathrm{Number}\text{of white balls}=\text{5}\\ \mathrm{Number}\text{of blue balls}=\text{5}\\ \mathrm{Number}\text{of ways to select 3 red balls}={\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{3}\\ =\frac{6!}{3!(6-3)!}\end{array}$

$\begin{array}{l}=\frac{6\times 5\times 4\times 3!}{3!.3!}\\ =20\\ \mathrm{Number}\text{of ways to select 3 white balls}={\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{3}\\ =\frac{5!}{3!(5-3)!}\\ =\frac{5\times 4\times 3!}{3!.2!}\\ =10\\ \mathrm{Number}\text{of ways to select 3 blue balls}={\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{3}\\ =\frac{5!}{3!(5-3)!}\\ =\frac{5\times 4\times 3!}{3!.2!}\\ =10\\ \mathrm{Number}\text{\hspace{0.17em}of ways of selecting 9 balls}=20\times 10\times 10\\ =2000\end{array}$

**Q.6** Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

**Ans.**

$\begin{array}{l}\mathrm{Number}\text{of cards in a deck}=52\\ \mathrm{Number}\text{of cards in each combination}=\text{5}\\ \text{Number of aces in a deck}=4\\ \text{Number of ace in each combination}=\text{1}\\ \text{Number of ways to select 1 ace out of 4}={\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{1}\end{array}$

$\begin{array}{l}=\frac{4!}{1!(4-1)!}\\ =\frac{4!}{1!3!}\\ =4\\ \text{\hspace{0.17em}No. of ways to select 4 cards out of 48}={\text{\hspace{0.17em}}}^{48}{\mathrm{C}}_{4}\\ =\frac{48!}{4!(48-4)!}\\ =\frac{48!}{4!\text{\hspace{0.17em}}44!}\\ =\frac{48\times 47\times 46\times 45\times 44!}{4\times 3\times 2\times 1\times 44!}\\ =2\times 47\times 46\times 45\\ =194580\\ \text{Number of 5 cards combination}=\text{\hspace{0.17em}}2\times 47\times 46\times 45\times 4\\ =778320\end{array}$

**Q.7** In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

**Ans.**

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Total}\text{ number of players}=17\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of players who can bowl}=\text{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Number of players in cricket team}=11\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of bowlers in cricket team}=4\end{array}$

$\begin{array}{l}\mathrm{Number}\text{\hspace{0.17em}of players who can not bowl}=17-5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12\\ \text{\hspace{0.17em}}\mathrm{Number}\text{ of ways to select 11 players}={\text{\hspace{0.17em}}}^{12}{\mathrm{C}}_{7}\times {\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{4}\\ =\frac{12!}{7!(12-7)!}\times \frac{5!}{4!(5-4)!}\\ =\frac{12!}{7!\text{\hspace{0.17em}}\times 5!}\times \frac{5!}{4!1!}\\ =\frac{12\times 11\times 10\times 9\times 8\times \overline{)7!}}{\overline{)7!\text{\hspace{0.17em}}}\times \overline{)5!}}\times \frac{\overline{)5!}}{4!}\\ =3960\end{array}$

**Q.8** A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

**Ans.**

$\begin{array}{l}\mathrm{Number}\text{of black balls in a bag}=5\\ \mathrm{Number}\text{of red balls in a bag}=6\\ \mathrm{Number}\text{of ways to select 2 black and 3 red balls}\\ ={\text{\hspace{0.33em}}}^{5}{\text{C}}_{2}\text{\hspace{0.33em}}\times {\text{\hspace{0.33em}}}^{6}{\text{C}}_{3}\\ =\frac{5!}{2!\left(5-2\right)!}\times \frac{6!}{3!\left(6-3\right)!}\\ =\frac{5!}{2!\text{\hspace{0.33em}}\times \text{\hspace{0.33em}}3!}\times \frac{6!}{3!\text{\hspace{0.33em}}\times \text{\hspace{0.33em}}3!}\\ =\frac{5\times 4\times \overline{)3!}}{2!\text{\hspace{0.33em}}\times \text{\hspace{0.33em}}\overline{)3!}}\times \frac{\overline{)6}\times 5\times 4\times \overline{)3!}}{\overline{)3!}\text{\hspace{0.33em}}\times \text{\hspace{0.33em}}\overline{)3!}}\\ =10\times 5\times 4\\ =200\end{array}$

**Q.9** In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

**Ans.**

$\begin{array}{l}\mathrm{Available}\text{courses for students}=9\\ \mathrm{Number}\text{of specific courses}=2\\ \mathrm{Courses}\text{to be chosen by students}=5\\ \mathrm{Not}\text{specific courses for students}=5-2\\ =3\\ \mathrm{Number}\text{of ways to select 3 remaining courses}\\ ={\text{\hspace{0.17em}\hspace{0.17em}}}^{7}{\mathrm{C}}_{3}\\ =\frac{7!}{3!\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}(7-3)!}[\because {\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}]\\ =\frac{7!}{3!\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}4!}\\ =\frac{7\times 6\times 5\times 4!}{3!\text{\hspace{0.17em}}\times \text{\hspace{0.17em}}4!}\\ =35\\ \mathrm{Thus},\text{a student can choose a programme of 5 courses in 35 ways.}\end{array}$

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