**NCERT solutions for class 11 maths exercise 7.1 of chapter 7-Permutations and Combinations**

NCERT solutions for class 11 maths exercise 7.1 of chapter 7-Permutations and Combinations are important for class 11 students to clear their concept about exercise 7.1 of chapter 7 Permutations and Combinations. These NCERT solutions of maths will help students in doing homework and their preparation of the exam, class assessments, and competitive entrance exams like polytechnic, engineering, NDA, and CDS . All solutions of unsolved questions of exercise 7.1 are created by an expert and scrutinized by a team of Future Study Point.

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**Exercise 2.2-Relations and Functions**

**Exercise 7.2 -Permutations and Combinations**

**Download -PDF of NCERT solutions for class 11 maths exercise 7.1 of chapter 7-Permutations and Combinations**

**PDF-NCERT solutions for class 11 maths exercise 7.1 of chapter 7-Pemutations and Combinations**

**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 |

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**NCERT solutions for class 11 maths exercise 7.1 of chapter 7-Pemutations and Combinations**

**Q1. How many three-digit numbers can be formed from the digits 1,2,3,4 and 5 assuming that**

**(i) Repetition of the digits is allowed?**

**(ii) Repetition of the digits is not allowed?**

Ans. (i) The given digits are 1,2,3,4 and 5

The number of three-digit numbers to be formed from the given digits = The number of ways of filling three places once, tense, and thousands of a three-digit number in succession by the given digits as follows.

The number of ways of filling unit place = 5

Since repetition of the digits allowed so, the number of ways of filling tense place= 5 and thousands place = 5

According to the multiplication rule total number of ways of filling three places =5×5×5 = 125

Therefore when repetition is allowed then 125 three-digit numbers can be formed from the given digits.

(ii) When repetition of the digits is not allowed, the number of ways of filling unit place = 5 ways

Number of ways of filling tense place by any of the remaining 4 digit numbers = 4 ways and number of ways of filling a hundred places by any of the remaining 3 digit numbers = 3 ways

According to the multiplication rule total number of ways of filling three places =5×4×3= 60

Therefore when repetition is not allowed then 60 three-digit numbers can be formed from the given digits.

**Q2. How many three-digit even numbers can be formed from the digits 1,2,3,4,5,6 if the digits can be repeated.**

Ans. (i) The given digits are 1,2,3,4,5 and 6

The number of three digit even numbers to be formed from the given digits = The number of ways of filling three places once, tense, and thousands of a three-digit number in succession by the given digits as follows.

Since the unit place of a three-digit even number should be any one of the even digits,2,4 or 6, therefore the number of ways of filling unit place = 3

Since repetition of the digits allowed so, the number of ways of filling tense place= 6 and hundred place = 6

According to the multiplication rule total number of ways of filling three places =3×6×6 = 108

Therefore when repetition is allowed then 108 three-digit even numbers can be formed from the given digits.

**Q3. How many four-letter codes can be formed using the first 10 letters of the English alphabet? If no letters can be repeated.**

Ans. The first 10 letters of the English alphabets are A, B, C, D, E, F, G, H, I, and J.

The number of four-letter codes formed = The number of ways of filling four vacant places of a code in succession by the given alphabets.

The number of ways of filling first place =10

Since repetition is not allowed, so the number of ways of filling second place = The number of remaining 9 letters =9

The number of ways of filling third place =The number of remaining 8 letters =8

The number of ways of filling fourth place =The number of remaining 7 letters =7

According to the multiplication rule total number of ways of filling four places of the code =10×9×8×7 = 5040

Therefore when repetition is not allowed then 5040 four letter codes can be formed from the given 10 alphabets.

**Q4.How many 5 digit telephone numbers can be constructed using the digits 0 and 9 if each number starts with 67 and no digits appear more than once?**

Ans. The given digits are 10 in number 0,1,2,3,4,5,6,7,8 and 9.

The number of 5 digit telephone numbers formed = The number of ways of filling 5 vacant places of a telephone number in succession by the given digits.

Since each telephone number starts with 67 , so first two places are fixed, then number of ways of filling third place by anyone remaining 8 digits= 8 ways

Since no digits appear more than once,so the number of ways of filling fourth place by the remaining 7 digits = 7 ways

The number of ways of filling fifth place by the remaining 6 digits = 6 ways

According to the multiplication rule total number of ways of filling five places of the telephone number =8×7×6 = 336

Therefore when repetition is not allowed then 336 ftelephone numbers can be formed from the given 10 digits.

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**Q5.A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?**

Ans. If a coin is tossed 3 times then outcomes are recorded as follows

The number of outcomes recorded in first time = 2 (head,tail)

The number of outcomes recorded in second time = 2 (head,tail)

The number of outcomes recorded in third time = 2 (head,tail)

According to the multiplication rule total number of possible outcomes = 2×2×2 = 8

**Q6. Given 5 flags of the different colours. How many different signals can be generated if each signal requires the use of 2 flags one below the other?**

Ans. The number of flags are = 5

Each signal requires the use of 2 flags one below the other

The number of different signals generated = The number of ways of filling two vacant places by anyone of 5 flags.

The number of ways of filling upper vacant place = 5

The number of ways of filling lower vacant place by the remaining four flags = 4

According to the multiplication rule total number of signals can be generated = 5×4 = 20

NCERT solutions for class 11 maths exercise 7.1 of chapter 7-Permutations and Combinations are based on the fundamental principle of counting which is a multiplication rule of total possibilities or outcome.

## The fundamental principle of counting

Let there m ways of doing a thing and n ways of doing another thing then total number of ways of doing both the things are m× n,this is also known as the multiplication rule.

**Example : **Let there are three pants P1,P2 and P3 and two shirts S1 and S2,the number of possible pairs out of three pants and two shirts would be as follows.

The number of ways of choosing three pants are = 3

The number of ways of choosing two shirts are = 2

Total number of possible pair are =3 × 2 = 6

It can be understood easily from the given figure.

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**NCERT Solutions for 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 |

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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 |

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Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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