# NCERT Solutions for Class 11 Maths Chapter 7

NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations (Kramchay aur Sanchay) Exercise 7.1, Exercise 7.2, Exercise 7.3, Exercise 7.4 and Miscellaneous download in PDF free for 2020-21.

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## NCERT Solutions for Class 11 Maths Chapter 7

 Class: 11 Maths (English and Hindi Medium) Chapter 7: Permutations and Combinations

### 11th Maths Chapter 7 Solutions

NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations all exercises including miscellaneous are given below for new academic session 2020-21. These NCERT solutions 2020-21 are for CBSE, Uttarakhand Board, Bihar Board, UP Board, MP Board, Gujrat Board and other state board’s students, who are following NCERT Books 2020-21 for their school exams 2020-21. Students can do directly class 12 without appearing in 11th through NIOS Online Admission. It is equally valuable as the other boards.

• ### 11th Maths Chapter 7 Solutions in PDF

#### Class 11 Maths Chapter 7 Exercise 7.1 Solution in Videos

Class 11 Maths Chapter 7 Exercise 7.1 Solution
Class 11 Maths Exercise 7.1 Solution in Hindi

#### Class 11 Maths Chapter 7 Exercise 7.2 Solution in Videos

Class 11 Maths Chapter 7 Exercise 7.2 Solution
Class 11 Maths Exercise 7.2 Solution in Hindi

#### Class 11 Maths Chapter 7 Exercise 7.3 Solution in Videos

Class 11 Maths Chapter 7 Exercise 7.3 Solution
Class 11 Maths Exercise 7.3 Solution in Hindi

#### Class 11 Maths Chapter 7 Exercise 7.4 Solution in Videos

Class 11 Maths Chapter 7 Exercise 7.4 Solution
Class 11 Maths Exercise 7.4 Solution in Hindi

#### Class 11 Maths Chapter 7 Miscellaneous Exercise 7 Solution in Videos

Class 11 Maths Chapter 7 Miscellaneous 7 Solution
Class 11 Maths 7 Miscellaneous 7 Solution in Hindi
##### What is meant by Permutation?

Permutation: A permutation is an arrangement of a number of objects in a definite order taken some or all at a time.

##### What is meant by Combination?

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).

#### Terms on Permutations & Combinations

1. 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.
2. 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)!)
3. 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.
4. 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).

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### Important Questions on 11th Maths Chapter 7

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?
When 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?
There will be as many ways as there are ways of filling 3 vacant places _ _ _ in succession by the given six digits. In this case, the units place can be filled by 2 or 4 or 6 only i.e., the units place can be filled in 3 ways. The tens place can be filled by any of the 6 digits in 6 different ways and also the hundreds place can be filled by any of the 6 digits in 6 different ways, as 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?
Each signal requires the use of 2 flags.
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! = 1 × 2 × 3 = 6 and 4! = 1 × 2 × 3 × 4 = 24
∴ 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?
There are 8 different letters in the word EQUATION.
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?
In the word EQUATION, there are 5 vowels, namely, A, E, I, O, and U, and 3 consonants, namely, Q, T, and N.
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?
5 men and 4 women are to be seated in a row such that the women occupy the even places.
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              