**NCERT Solutions for Class 11 Maths Exercise 7.4 of the Chapter 7 Permutaion and Combination**

NCERT Solutions for Class 11 Maths Exercise 7.4 of the Chapter 7 Permutaion and Combination are available here for boosting the preparation of class 11 maths exam. The NCERT Solutions for Class 11 Maths Exercise 7.4 of the Chapter 7 Permutaion and Combination are created by an expert of maths who has huge experience in teaching maths from class 9 to class 12.All the questions of permutation and combination are solved by step by step ways therefore every students of the grade 11 can understand the solutions easily .

**NCERT Solutions for Class 11 Maths Chapter 7 Permutation and Combination**

**Exercise 7.1- Permutations and Combinations**

**Exercise 7.2 -Permutation and Combination**

**Exercise 7.3-Permutation and Combination**

**Exercise 7.4-Permutation and Combination**

**Miscellaneous Exercise-Permutation and Combination**

## NCERT Solutions for Class 11 Maths Exercise 7.4 of the Chapter 7 Permutaion and Combination

**Q1. If ^{n}C_{8} = ^{n}c_{2}, find ^{n}c_{2}**

Ans. Since ^{n}C_{r }=^{n}C_{n-r}

∴^{n}C_{8 }= ^{n}C_{n-8, }

Substituting ^{n}C_{8 }= ^{n}C_{n-8,}

^{n}C_{n-8} = ^{n}c_{2}

n – 8 = 2⇒ n = 10

Hence the required value of ^{n}c_{2}is 45

**Q2.Determine n if**

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

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

Ans. It is given to us that

^{2n}C_{3}: ^{n}c_{3 }= 12 : 1

4(2n -1)(n -1) : (n -1)(n -2) = 12 : 1

8n -4 = 12n – 24

4n = 20

n = 5

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

Similarly as in (i),we have

11n – 22 = 8n – 4

3n = 18

n = 6

**Q3. How many chords can be drawn through 21 points on a circle?**

Ans.For drawing chords we are required to choose two points out of 21 points on the circle, therefore the number of ways of such combinations are ^{21}C_{2}

The number of chords are = ^{21}C_{2}

Hence total number of chords can be drawn are 210

**Q4.In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls.**

Ans. The number of ways 3 boys can be selected out of 5 boys are ^{5}C_{3}

The number of ways 3 girls can be selected out of 4 girls are ^{4}C_{3}

According to the multiplication rule,total number of ways selecting a team of 3 boys and 3 girls are = ^{5}C_{3 }×^{4}C_{3}

Hence the number of ways a team of 3 boys and 3 girls selected from 5 boys and 4 girls is 40

**Q5.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. The number of ways of selecting 3 balls out of 6 red balls are ^{6}C_{3}

The number of ways of selecting 3 balls out of 5 white balls are ^{5}C_{3}

The number of ways of selecting 3 balls out of 5 blue balls are ^{5}C_{3}

According to the multiplication rule,total number of ways selecting 9 balls out of 6 red ball,5 white balls and 5 blue balls are = ^{6}C_{3 }×^{5}C_{3 }×^{5}C_{3}

= 20 × 10 × 10 = 2000

Hence required number of ways of selecting 9 balls are 2000

**Q6.Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.**

Ans. The number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination are =The number of ways selecting 1 ace out of 5 cards × The number of ways selecting remaining 4 cards out of remaining (52 -4= 48 ) cards

The number of ways selecting 1 ace out of 4 cards = ^{4}C_{1}

The number of ways selecting remaining 4 cards out of remaining (52 -4 = 48) cards = ^{48}C_{4}

The number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination are = ^{4}C_{1 }× ^{48}C_{4}

= 4× 2× 47 ×46 ×45 = 778320

The number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination is 778320

**Q7.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.The number of ways selecting a team of 11 players = The number of ways selecting 4 bowlers out of 5 bowlers × The number of ways selecting 7 players out of remaining (17-5 =12) players

The number of ways selecting a team of 11 players = ^{5}C_{4}×^{12}C_{7}

=5 × 11×9×8 = 3960

The number of ways selecting a team of 11 players in which there are exactly 4 bowlers is 3960

**Q8.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.The number of ways selecting 2 black balls out of 5 black balls = ^{5}C_{2}

The number of ways selecting 3 red balls out of 6 red balls = ^{6}C_{3}

The number of ways selecting 2 black balls and 3 red balls out of 5 black balls and 6 red balls =^{5}C_{2 }× ^{6}C_{3}

= 10 × 20 = 200

The number of ways selecting 2 black balls and 3 red balls out of 5 black balls and 6 red balls is 200

**Q9.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. It is given that student has to choose 5 coarses out of 9 coarses in which 2 coarses are compulsury.

Since 2 coarses are compusury therefore number of ways selecting 5 coarses = Number of ways of selecting remaining (5 -2 = 3) coarses out of remaining (9 -2 = 7) coarses is = ^{7}C_{3}

Hence required number of ways selecting 5 coarses is 35

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**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

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Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

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Chapter 6-Lines and Angles | Chapter 14-Statistics |

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**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions of class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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