**NCERT solutions for class 11 maths exercise 7.2 of chapter 7-Permutations and Combinations**

**Download PDF of NCERT solutions for class 11 maths exercise 7.2 of chapter 7-Permutations and Combinations**

NCERT solutions for class 11 maths exercise 7.2 of chapter 7-Permutations and Combinations are prepared for helping the students in their maths study so that they could clear their maths concepts.NCERT solutions for class 11 maths exercise 7.2 of chapter 7-Permutations and Combinations are based on permutations.A permutation is the number of arrangements formed by choosing a certain number of objects at a time from a set of objects.

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**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 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**

**Study notes of Maths and Science NCERT and CBSE from class 9 to 12**

**Q1. Evaluate **

**(i) 8 ! (ii) 4 ! – 3 !**

Ans.(i) 8 ! = 8 ×7 ×6 ×5 ×4 ×3 ×2 ×1 = 40320

(i) 4 !- 3! = 4 ×3 ×2 ×1-3×2×1= 24-6 = 18

**Q2. Is 3 ! + 4 ! = 7! ? **

Ans. Taking LHS

3 ! + 4 ! =3×2×1 +4× 3×2×1 = 6 +24 =30

RHS= 7! = 7×6×5 ×4× 3×2×1=5040

∴ 3 ! + 4 ! ≠ 7!

**Q3. Compute :**

Ans.

Ans.

x = 64

**(i) n = 6,r= 2 (ii) n = 9, r = 5**

Ans.(i) n = 6,r= 2

** (ii) n = 9, r = 5**

Exercise 7.3

**Q1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?**

Ans. The given digits are 9, i.e 1,2,3….9

The number of 3 digits numbers using 9 digits = Permutations of 9 different digits choosing 3 digits at a time

Hence required number of 3 digit numbers

= ^{9}P_{3}

**Q2. How many 4-digit numbers are there with no digit repeated?**

Ans. The digits utilized to form 4 digits numbers are 0,1,2,3….9

Number of 4-digit numbers with no digit repeated = Number of permutations of 9 different digits choosing 4 digits at a time

Hence required number of 4 digit numbers

= ^{10}P_{4}

Since the number of 5040 also includes the numbers which starts from 0 i.e 0123,0324 etc,but actually these are three digits numbers

Therefore the number of permutations of 9 different digits (o is excluded) choosing 3 digits at a time

= ^{9}P_{4}

Therefore the required 4 digit numbers are = 5040 – 504 = 3536

**Q3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6,7 if no digit is repeated.**

Ans. Let’s find out 3-digit even numbers using the digits 1, 2, 3, 4,6,7

For a three-digit even number , any of the 3 digits 2,4, or 6 is required in its unit place.

The number of ways of filling unit place is = The number of permutations of 3 different digits choosing 1 digits (2,4,or 6) at a time

= ^{3}P_{1}

Since digits are not allowed to be repeated then the number of ways of filling tens and hundred places are = Number of permutations of remaining 5 digits choosing 2 digits at a time

= ^{5}P_{2}

According to the multiplication principle the number of ways creating three-digit even numbers are = 3 × 20 = 60

Therefore required 3 digit even numbers are = 60

**Q4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?**

Ans.The number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated = Number of permutations of 5 digits choosing 4 digits at a time

= ^{5}P_{4}

Let’s find out 4-digit even numbers using the digits 1, 2, 3, 4,5

For a four-digit even number, any of the 2 digits 2,4 is required in its unit place.

The number of ways of filling unit place is = The number of permutations of 2 different digits choosing 1 digits (2,4,) at a time

= ^{2}P_{1}

Since digits are not allowed to be repeated then the number of ways of filling tens and hundred and thousand places are = Number of permutations of remaining 4 digits choosing 3 digits at a time

= ^{4}P_{3}

According to the multiplication principle the number of ways of creating four-digit even numbers are = 2 × 24 =48

Therefore required 4 digit even numbers are 48 and number of 4-digit numbers that can be formed are 120.

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**Q5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice-chairman assuming one person can not hold more than one position?**

Ans. From a committee of 8 persons, let’s find out the number of ways of choosing a chairman and a vice-chairman assuming one person can not hold more than one position

The number of ways of choosing a chairman and a vice-chairman out of a committee of 8 persons = The number of permutations of 8 different persons choosing 2 persons at a time

= ^{8}P_{2}

**Q6. Find n if ^{n-1}P_{3 }: ^{n}P_{4 }= 1 : 9**

Ans. ^{n-1}P_{3 }: ^{n}P_{4 }= 1 : 9

n = 9

**Q7. Find r if (i) ^{5}P_{r }= 2.^{6}P_{r-1 }**

**(ii)**

^{5}P_{r }=^{6}P_{r-1 }**Ans.**

**(i) ^{5}P_{r }= 2.^{6}P_{r-1 }**

(7 -r)(6 -r) = 12

42 – 7r – 6r + r² = 12

r²- 13r + 30 = 0

r² – 10 r – 3r + 30 = 0

r( r- 10) – 3(r – 10) =0

( r- 10)(r – 3) = 0

r = 10,3

In ^{n}P_{r },0≤r ≤n, therefore the value of r is 3

_{ }**(ii) ^{5}P_{r }= ^{6}P_{r-1 }**

(7 – r)(6 – r) = 6

42 – 7r – 6r + r² = 6

r² – 13r + 36 = 0

r² – 9r – 4r – 36 = 0

r(r – 9) – 4(r – 9) = 0

(r – 9) (r – 4) = 0

r =9,4

In ^{n}P_{r },0≤r ≤n, therefore the value of r is 4

**Q8. How many words can be, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter once?**

Ans. The number of letters in the word EQUATION are = 8

The number of ways,the words with or without meaning, can be formed using all 8 letters = Number of permutations of 8 letters choosing 8 letters at a time

= ^{8}P_{8}

**Q9. How many words can be, with or without meaning, can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.**

**(i) 4 letters are used at a time**

**(ii) all letters are used at a time**

**(iii) all letters are used but first letter is vowel?**

Ans.(i) The number of letters in the word MONDAY are = 6

The number of ways,the words with or without meaning, can be formed using 4 letters at a time out of 6 letters= Number of permutations of 6 letters choosing 4 letters at a time

= ^{6}P_{4}

_{}

(ii) The number of ways,the words with or without meaning, can be formed using all 6 letters at a time out of 6 letters= Number of permutations of 6 letters choosing 6 letters at a time

= ^{6}P_{6}

(iii) The number of vowels in the word MONDAY is = 2 (i.e O, A)

The number of ways the words formed using 1 vowel at a first place out of 2 vowels= Number of permutations of 2 vowels choosing 1 vowel at a time

= ^{2}P_{1 }= 2

Since one place is already occupied by one vowel therefore the number of ways,of selecting the remaining 5 letters at a time = Number of permutations of 5 letters choosing 5 letters at a time

= ^{5}P_{5 }= 5! = 5×4×3×2= 120

According to the multiplication principle, the number of ways of creating 6 letters words are = 2 × 120 =240

**Q10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?**

Ans. The total number of words in MISSISSIPPI are = 11

The frequency of I = 4

The frequency of S = 4

The frequency of P = 2

We have to determine

Distinct permutations of the letters in word when 4I’s not come together= Distinct permutations of the letters in the word -Distinct permutations of the letters in word when 4I’s come together

Number of words formed by using all the letters of the word =The distinct permutations of the letters in the word

When 4I’s come together should be treated as a single object, therefore in this case total letters would be supposed as 8, with the frequency of S =4 and P=2

The distinct permutations of the letters in the word when 4 I’s are together

Distinct permutations of the letters in word when 4I’s not come together =34650-840= 33810

**Q11. In how many ways can the letters of the word PERMUTATIONS be arranged if the**

**(i) words start with P and end with S**

**(ii) vowels are all together**

**(iii) there are always 4 letters between P and S?**

Ans. (i) The number of letters in word PERMUTATIONS are = 12

Let’s find out the words start with P and end with S, means first and last letter are fixed, permutations of letters of the remaining 10 letters in which there are 2 T ‘s

Therefore in this case the number of ways of arrangement

**(ii) **In this case when vowels are all together,so all vowels together are treated as single letter.There are 5 vowels A,E,I,O,U, reamaining 7 letters and 5 vowels treated as 8 letters with 2 T’s

Therefore in this case the number of ways of arrangement

The number of ways of arrangement of 5 vowels = 5! = 5×4×3×2=120

Hence according to multiplication principle total number of arrangements in this case =20160 × 120 = 2419200

(iii) There are always 4 letters between P and S, it can be shown as below.

It can be observed that the number of ways of placing 4 letters between P and S are = 7 ways

The way of arrangement of P and S is =2! = 2

The remaining 10 letters can be arranged with 2 I’s by (10!)/2!

According to the multiplication principle, the number of ways of placing 4 letters between P and S are = 7×2 × (10!)/2! =25401600

**See the video of the solutions of question number 11**

## Permutations

A permutation is the number of arrangements formed by choosing a certain number of objects at a time from a set of objects.Let there are n objects and we have to creat the arrangement of the objects with r objects taken at a time then permutations of the n objects r taken at a time

If certain objects repeated by m and t times then permutations of all objects taken at a time

Permutations are the way of arrangement of the things when order also matters, which means in the arrangement if we have taken 23 then 32 is also counted.

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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 |

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

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

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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 |

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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