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NCERT Solutions for Class 11 Maths Chapter 6 Permutations And Combinations Ex 6.3

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Get NCERT Solutions for Maths Class 11 Chapter 6 Exercise 6.3 - FREE PDF Download

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations, Exercise 6.3, involve arranging objects in specific orders. This topic is essential for understanding the different ways to order elements, with applications in probability and other mathematical areas. The focus is on learning the formulas for permutations and solving various problems to apply these concepts effectively.

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Students should concentrate on the fundamental principles, practice different types of permutation problems, and understand the importance of order in arrangements in Exercise 6.3 Class 11 Maths NCERT Solutions. This chapter lays the groundwork for more advanced mathematical topics and enhances problem-solving skills. Access CBSE Class 11 Maths Syllabus for the complete Maths solutions for Class 11 as per the updated Syllabus.


Formulas Used in Class 11 Chapter 6 Exercise 6.3

  • nPr=n!(nr)!

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Access NCERT Solutions for Maths Class 11 Chapter 6 - Permutations and Combinations

Exercise 6.3

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

Ans:

We have to form a 3-digit numbers using digits 19. That means, the order of the digits matter.

Hence, there will be as many 3-digit numbers as there are permutations of 9 digits taken 3 at a time.

So, required number of 3-digit numbers 

= 9P3

 9!(93)!=9!6!

9×8×7×6!6!

9×8×7=504

Hence, there are 504  3-digit numbers formed.


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

Ans:

The thousands place of the 4-digit number is to be filled with any of the digits from 1 to 9 as the digit 0 cannot be included. Therefore, the number of ways in which thousands place can be filled is 9. The hundreds, tens, and units place can be filled by any of the digits from 0 to 9. However, the digits cannot be repeated in the 4-digit numbers and thousands place is already occupied with a digit. The hundreds, tens, and units place is to be filled by the remaining 9-digits. Therefore, there will be as many such 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.

Number of such 3-digit numbers = 9P3

9!(93)!=9!6!

9×8×7×6!6!

9×8×7=504

Thus, by multiplication principle, the required number of 4-digit numbers are 9×504=4536.


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

Ans:

We have to form 3-digit even numbers using the given six digits, 1, 2, 3, 4, 6, ,and 7, without repeating the digits. Then, units can be filled in 3 ways by any of the digits, 2, 4 or 6. Since the digits cannot be repeated in the 3-digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining 5-digits. Therefore, the number of ways in which hundreds and tens place can be filled with the remaining 5-digits is the permutation of 5 different digits taken 2 at a time.

So, the number of ways of filling hundreds and tens place

5P2

5!(52)!=5!3!

5×4×3!3!

5×4=20

Thus, by multiplication principle, the required number of 3-digit numbers are 3×20=60.


4. 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:

We have to form 4-digit numbers using the digits, 1, 2, 3, 4, and 5

So, there will be as many 4-digit numbers as there are permutations of 5 different digits taken 4 at a time.

That is, required number of 4-digit numbers

5P4

5!(54)!=5!1!

1×2×3×4×5

120

Among the 4-digit numbers formed by using the digits, 1, 2, 3, 4, 5 even numbers end with either 2 or 4. The number of ways in which units place is filled with digits is 2. Since the digits are not repeated and the units place is already occupied with a digit (which is even), the remaining places are to be filled by the remaining 4-digits. Therefore, the number of ways in which the remaining places can be filled is the permutation of 4 different digits taken 3 at time. 

Number of ways of filling the remaining places

4P3=4!(43)!

4!

4×3×2×1=24

Thus, by multiplication principle, the required number of even numbers is 24×2=48.

                                                                               

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

Ans:

We have to choose a chairman and a vice chairman from a committee of 8 persons in such a way that one person cannot hold more than one position. Here, the number of ways of choosing a chairman and a vice chairman is the permutation of 8 different objects taken 2 at a time. 

Thus, required number of ways

8P2=8!(82)!

8×7×6!6!=8×7

56

6. Find n if n1P3:nP4=1:9.

Ans:

Here,

n1P3:nP4=1:9

n1P3nP4=19

[(n1)!(n13)!][n!(n4)!]=19

(n1)!(n4)!×(n4)!n!=19

(n1)!n×(n1)!=19

1n=19

Hence, n=9.

7. Find r if:

(i). 5Pr=26Pr1

Ans:

5!(5r)!=2×6!(6r+1)!

5!(5r)!=2×6×5!(7r)(6r)(5r)!

(7r)(6r)=12

426r7r+r2=12

r(r3)10(r3)=0

r=3 or r=10

It is known that nPr=n!(nr)!, where 0rn.

0r5

Hence, r=3.

(ii). 5Pr=6Pr1

Ans: 

5!(5r)!=6!(6r+1)!

5!(5r)!=6×5!(7r)(6r)(5r)!

(7r)(6r)=6

426r7r+r2=6

r(r4)9(r4)=0

It is known that nPr=n!(nr)!, where 0rn.

0r5

Hence, r=4.


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

Ans:

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 8P8=8!.

Thus, required number of words that can be formed is

8!=40320


9. How many words, 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,

Ans:

There are 6 different letters in the word MONDAY.

Number of 4-letter words that can be formed from the letters of the word MONDAY, without repetition of letters, is the number of permutations of 6 different objects taken 4 at a time, which is 6P4.

Thus, required number of words that can be formed using 4 letters at a time are 6P4=6!(64)!

6×5×4×3×2!2!=6×5×4×3

360

(ii). all letters are used at a time,

Ans:

Number of words that can be formed by using all the letters of the word MONDAY at a time is the number of permutations of 6 different objects taken 6 at a time, which is 6P6=6!.

Thus, required number of words that can be formed when all letters are used at a time

6!=1×2×3×4×5×6.

720

(iii). all letters are used but first letter is a vowel.

Ans:

In the given word, there are 2 different vowels, which have to occupy the rightmost place of the words formed. This can be done only in 2 ways. Since the letters cannot be repeated and the rightmost place is already occupied with a letter (which is a vowel), the remaining five places are to be filled by the remaining 5 letters. This can be done in 5! ways.

Thus, in this case, required number of words that can be formed is 5!×2=120×2

240


10. In how many of the distinct permutations of the letters in MISSISSIPP do the four I’s not come together?

Ans: 

In the given word MISSISSIPPI, I appear 4 times, S appears 4 times, P appears 2 times, and M appears just once. Therefore, number of distinct permutations of the letters in the given word

11!4!4!2!

11×10×9×8×7×6×5×4!4!×4×3×2×1×2×1

11×10×9×8×7×6×54×3×2×1×2×1

34650

There are 4 I’s in the given word. When they occur together, they are treated as single object [IIII] for the time being. This single object together with the remaining 7 objects will account for 8 objects.

These 8 objects in which there are 4 S’s, and 2 P’s can be arranged in 8!4!2! ways that means, 840 ways.

Number of arrangements where all I’s occurred together is in 840 ways.

Thus, number of distinct permutations of the letters in MISSISSIPPI in which four IS do not come together 34650840=33810.


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

(i). Words start with P and end with S,

Ans:

In the word PERMUTATIONS, there are 2 T’s, and all the other letters appear only once.

If P and S are fixed at the extreme ends (P at the left end and S at the right end), then 10 letters are left.

Hence, in this case, required number of arrangements

10!2!=1814400.

(ii). Vowels are all together,

Ans:

There are 5 vowels in the given word, each appearing only once.

Since they have to always occur together, they are treated as a single object for the time being. This single object together with the remaining 7 objects will account for 8 objects, in total. 

These 8 objects in which there are 2 T’s can be arranged in 8!2! ways.

Corresponding to each of these arrangements, the 5 different vowels can be arranged in 5! ways. Therefore, by multiplication principle, required number of arrangements in this case

8!2!×5!=2419200.

(iii). There are always 4 letters between P and S?

Ans:

The letters have to be arranged in such a way that there are always 4 letters between P and S. Therefore, in a way, the places of P and S are fixed. The remaining 10 letters in which there are 2 T’s can be arranged in 10!2! ways.

Since,Possible places of P & S are 1 & 6, 2 & 7, 3 & 8, 4 & 9, 5 & 10, 6 & 11, 7 & 12. Then there can be 4 letters between P & S.Also P & S can be interchanged as it won’t effect the number of letters between them. 

So, the letters P and S can be placed such that there are 4 letters between them in 2×7=14 ways.Therefore, by multiplication principle, required number of arrangements in this case

10!2!×14=25401600.

Conclusion

In conclusion, Chapter 6 permutations and combinations Ex 6.3 class 11 maths NCERT solutions contains key concepts, emphasizing the importance of order and selection. Focus on understanding the differences between the two, mastering formulas, and practising diverse problems. These topics not only enhance problem-solving skills but also prepare students for more advanced studies in mathematics. Previous year question papers typically feature 3–4 questions on these topics, highlighting their significance in exams. Consistent practice and a clear grasp of concepts are essential for success in this area. 


Class 11 Maths Chapter 6: Exercises Breakdown

Exercise

Number of Questions

Exercise 6.1

6 Questions and Solutions

Exercise 6.2

5 Questions and Solutions

Exercise 6.4

9 Questions and Solutions

Miscellaneous Exercise

11 Questions and Solutions


CBSE Class 11 Maths Chapter 6 Other Study Materials


Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Important Related Links for CBSE Class 11 Maths

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FAQs on NCERT Solutions for Class 11 Maths Chapter 6 Permutations And Combinations Ex 6.3

1. What is a permutation studied in exercise 6.3 class 11 permutations and combinations?

A permutation is an arrangement of objects in a specific order, where the sequence matters. Permutations are a foundational concept that enhances problem-solving and analytical skills in mathematics.

2. How is permutation different from combination taught in exercise 6.3 class 11 maths?

In permutations, order is important. In combinations, order does not matter.

3. What is the formula for permutations in chapter 6 maths class 11 exercise 6.3?

The formula for permutations of n objects taken r at a time is nPr=n!(nr)!.

4. What does n! (n factorial) mean in NCERT Class 11 Maths Chapter 6 Exercise 6.3 ?

n! (n factorial) is the product of all positive integers up to n.

5. How do you calculate permutations with identical objects mentioned in ex 6.3 class 11 permutations and combinations?

Divide by the factorial of the number of identical objects to avoid overcounting.

6. What are circular permutations in exercise 6.3 class 11 math?

Circular permutations are arrangements of objects in a circle, where rotations of the same arrangement are considered identical.

7. What are permutations with repetition in exercise 6.3 class 11 math?

Permutations with repetition allow repeated use of the same element.

8. Why are permutations important in NCERT class 11 maths chapter 6 exercise 6.3?

Permutations help in solving problems where order is crucial, such as seating arrangements and password generation.

9. How many questions on permutations were asked in previous year exams?

Typically, 2–3 questions on permutations appear in previous year question papers.

10. What should students focus on while studying permutations in NCERT solutions for class 11 maths chapter 6 exercise 6.3?

Focus on understanding the formulas, practising diverse problems, and grasping the concept of order in arrangements.