NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations (क्रमचय और संचय) Exercise 7.1 or Exercise 7.2 or Exercise 7.3 or Exercise 7.4 or Miscellaneous view online or download in PDF form free.

Class: | 11 |

Subject: | Maths |

Chapter 7: | Permutations and Combinations |

Table of Contents

- 1 NCERT Solutions for Class 11 Maths Chapter 7
- 1.1 Class 11 Maths Chapter 7 Permutations & Combinations Solutions
- 1.1.1 Terms on Permutations & Combinations
- 1.1.2 How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is not allowed?
- 1.1.3 How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
- 1.1.4 Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
- 1.1.5 Is 3! + 4! = 7!?
- 1.1.6 How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
- 1.1.7 How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
- 1.1.8 It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

- 1.1 Class 11 Maths Chapter 7 Permutations & Combinations Solutions

## NCERT Solutions for Class 11 Maths Chapter 7

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### Class 11 Maths Chapter 7 Permutations & Combinations Solutions

These solutions are for CBSE, Uttarakhand Board, Bihar Board, UP Board, MP Board, Gujrat Board and other state board’s students, who are following NCERT Books for their school exams 2019-20. Students can do directly class 12 without appearing in 11th through NIOS Online Admission. It is equally valuable as the other boards.

- View Online Exercise 7.1 Sols
- Download Exercise 7.1 in PDF
- View Online Exercise 7.2 Sols
- Download Exercise 7.2 in PDF
- View Online Exercise 7.3 Sols
- Download Exercise 7.3 in PDF
- View Online Exercise 7.4 Sols
- Download Exercise 7.4 in PDF
- View Online Miscellaneous Exercise
- Download Miscellaneous Exercise 7
- NCERT Books for Class 11
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- Hindi Medium Solutions will be uploaded very soon.

##### 11 Maths Chapter 7 Exercise 7.1 Sols

NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Exercise 7.1 is given below. For other questions, please visit to Exercise 7.2 or Exercise 7.3 or Exercise 7.4 or Miscellaneous Solutions. Visit to Class 11 Maths main page or Top of the page.

##### 11 Maths Chapter 7 Exercise 7.2 Sols

NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.2 is given below. For other questions, please visit to Exercise 7.1 or Exercise 7.3 or Exercise 7.4 or Miscellaneous Solutions. Visit to Class 11 Maths main page or Top of the page.

##### 11 Maths Chapter 7 Exercise 7.3 Sols

Class 11 Maths Chapter 7 Exercise 7.3 solutions are given below. For other questions, please visit to Exercise 7.1 or Exercise 7.2 or Exercise 7.4 or Miscellaneous Solutions. Visit to Class 11 Maths main page or Top of the page.

##### 11 Maths Chapter 7 Exercise 7.4 Sols

Solutions for Class 11 Maths Chapter 7 Exercise 7.4 is given below. For other questions, please visit to Exercise 7.1 or Exercise 7.2 or Exercise 7.3 or Miscellaneous Solutions. Visit to Class 11 Maths main page or Top of the page.

##### 11 Maths Chapter 7 Miscellaneous Exercise

NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Miscellaneous Exercise is given below. For other questions, please visit to Exercise 7.1 or Exercise 7.2 or Exercise 7.3 or Exercise 7.4 Solutions. Visit to Class 11 Maths main page or Top of the page.

Visit to Class 11 Maths main page or Top of the page

#### Terms on Permutations & Combinations

*Permutation*: A permutation is an arrangement of a number of objects in a definite order taken some or all at a time.*Combination*: Each of the different selections made by choosing some or all of a number of objects, without considering their order is called a combination. The number of combination of n objects taken r at a time where 0 ≤ r ≤ n, is denoted by nCr or C(n, r).

*Multiplication Principle*(Fundamental Principle of Counting): If an event can occur in m different ways, following which another event can occur in n different ways, then the total no. of different ways of occurrence of the two events in order is m × n.*Factorial*: Factorial of a natural number n, denoted by n! or n is the continued product of first n natural numbers. n! = n × (n – 1) × (n – 2) × … × 3 × 2 × 1 or equal to n × ((n – 1)!) or equal to n × (n – 1) × ((n – 2)!)*Fundamental Principle of Addition*: If there are two events such that they can occur independently in m and n different ways respectively, then either of the two events can occur in (m + n) ways.

- The number of permutation of n different objects taken r at a time where 0 ≤ r ≤ n and the objects do not repeat is denoted by nPr or P(n, r).

Table of Contents

- 1 How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is not allowed?
- 2 How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
- 3 Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
- 4 Is 3! + 4! = 7!?
- 5 How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
- 6 How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
- 7 It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

#### How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is not allowed?

In this case, repetition of digits is not allowed. Here, if units place is filled in first, then it can be filled by any of the given five digits. Therefore, the number of ways of filling the units place of the three-digit number is 5.

Then, the tens place can be filled with any of the remaining four digits and the hundreds place can be filled with any of the remaining three digits.

Hence, by the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 × 4 × 3 = 60

#### How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Therefore, by multiplication principle, the required number of three digit even numbers is 3 × 6 × 6 = 108

#### Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

There will be as many flags as there are ways of filling in 2 vacant places _ _ in succession by the given 5 flags of different colours.

The upper vacant place can be filled in 5 different ways by any one of the 5 flags following which, the lower vacant place can be filled in 4 different ways by any one of the remaining 4 different flags.

Thus, by multiplication principle, the number of different signals that can be generated is 5 × 4 = 20

#### Is 3! + 4! = 7!?

∴ 3! + 4! = 6 + 24 = 30

But 7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040

Hence, 3! + 4! ≠ 7!

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

Therefore, the number of words that can be formed using all the letters of the word EQUATION, using each letter exactly once, is the number of permutations of 8 different objects taken 8 at a time,

Which is P (8, 8) =8! .

Hence, required number of words that can be formed = 8! = 40320

#### How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. Then, the permutations of these 2 objects taken all at a time are counted.

This number would be P(2, 2)=2!

Corresponding to each of these permutations, there are 5! permutations of the five vowels taken all at a time and 3! permutations of the 3 consonants taken all at a time.

Hence, by multiplication principle, required number of words = 2! × 5! × 3! = 1440

#### It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

The 5 men can be seated in 5! ways.

The 4 women can be seated only at the cross marked places (so that women occupy the even places).

M×M×M×M×M

Therefore, the women can be seated in 4! ways.

Hence, possible number of arrangements = 4! × 5! = 24 × 120 = 2880

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