NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.2 Permutations and Combinations help students understand the concept of permutations, which deals with the arrangement of objects where the order matters. This exercise builds on basic counting principles and introduces standard permutation formulas.
Prepared according to the latest CBSE Class 11 Maths syllabus, Exercise 6.2 focuses on calculating permutations using formulas and solving arrangement-based problems. Regular practice of this exercise helps students improve logical thinking and problem-solving skills.
NCERT Solutions For Class 11 Maths Chapter 6 Permutations And Combinations Exercise 6.2
Q.
Evaluate: (i) 8! (ii) 4! – 3!
Q.
NCERT Solutions For Class 11 Maths Chapter 6 Permutations And Combinations Exercise 6.2
The solutions are explained in a clear, step-by-step format so students can easily apply the correct formulas and avoid mistakes in exams.
1. Evaluate.
(i) 8!
Ans: A number's factorial is the function that multiplies it by each natural number below it. The product of the first n natural numbers is n factorial, which is written as n! and is given by,
n! = n × (n − 1)!
Now, the given expression is 8!
According to the definition of the factorial,
⇒ 8! = 8 × (8 − 1)!
⇒ 8! = 8 × 7!
This is further evaluated as,
⇒ 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
⇒ 8! = 40320
Therefore, the value of 8! is found to be 40320.
(ii) 4! − 3!
Ans: The product of the first n natural numbers is n factorial, which is written as n! and is given by,
n! = n × (n − 1)!
Now, the given expression is 4! − 3!
First, consider and evaluate 4!
According to the definition of the factorial,
⇒ 4! = 4 × (4 − 1)!
⇒ 4! = 4 × 3!
This is further evaluated as,
⇒ 4! = 4 × 3 × 2 × 1
⇒ 4! = 24
Thus, the value of 4! is found to be 24.
Now, consider and evaluate for 3!
According to the definition of the factorial,
⇒ 3! = 3 × (3 − 1)!
⇒ 3! = 3 × 2!
This is further evaluated as,
⇒ 3! = 3 × 2 × 1
⇒ 3! = 6
Thus, the value of 3! is found to be 6.
Finally, consider the expression, 4! − 3! and evaluate.
⇒ 4! − 3! = 24 − 6
⇒ 4! − 3! = 18
Therefore, the value of the expression on evaluation is found to be 18.
2. Is 4! + 3! = 7!?
Ans: The product of the first n natural numbers is n factorial, which is written as n! and is given by,
n! = n × (n − 1)!
First, consider and evaluate 4!
According to the definition of the factorial,
⇒ 4! = 4 × (4 − 1)!
⇒ 4! = 4 × 3!
This is further evaluated as,
⇒ 4! = 4 × 3 × 2 × 1
⇒ 4! = 24
Thus, the value of 4! is found to be 24.
Now, consider and evaluate for 3!
According to the definition of the factorial,
⇒ 3! = 3 × (3 − 1)!
⇒ 3! = 3 × 2!
This is further evaluated as,
⇒ 3! = 3 × 2 × 1
⇒ 3! = 6
Thus, the value of 3! is found to be 6.
Now, consider the expression, 4! + 3! and evaluate.
⇒ 4! + 3! = 24 + 6
⇒ 4! + 3! = 30
Further, consider 7!
According to the definition of the factorial,
⇒ 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
⇒ 7! = 5040
Therefore, it is observed that 4! + 3! ≠ 7!
3. Compute
8! / 6! × 2!
Ans: The product of the first n natural numbers is n factorial, which is written as n! and is given by,
n! = n × (n − 1)!
First consider the expression in the numerator, that is, 8!
According to the definition of the factorial,
⇒ 8! = 8 × (8 − 1)!
⇒ 8! = 8 × 7!
This is further evaluated as,
⇒ 8! = 8 × 7 × 6!
Now, consider the expression in the denominator, that is, 6! × 2!
This expression can also be written in way which is,
6! × 2 × 1
Consider the expression that is to be determined and evaluated.
⇒ 8! / 6! × 2! = (8 × 7 × 6!) / (6! × 2 × 1)
⇒ 8! / 6! × 2! = (8 × 7) / 2
⇒ 8! / 6! × 2! = 28
Therefore, the value of the expression given after evaluation is found to be 28.
4.
1/6! + 1/7! = x/8!
Ans: The product of the first n natural numbers is n factorial, which is written as n! and is given by,
n! = n × (n − 1)!
The given expression is
1/6! + 1/7! = x/8!
Consider making the denominator of the left hand side of the given expression equal.
⇒ 1/6! + 1/(7 × 6!) = x/8!
Take out the common terms aside,
⇒ 1/6! (1 + 1/7) = x/8!
⇒ 1/6! (8/7) = x/8!
Now, rearrange the above equation and solve for x.
⇒ x = 8!/6! (8/7)
⇒ x = (8 × 8 × 7 × 6!) / (7 × 6!)
⇒ x = 8 × 8
⇒ x = 64
Therefore, the value of x for the expression on evaluation is found to be 64.
5. Evaluate n! / (n − r)!, when
(i) n = 6, r = 2
Ans: The expression given is
n! / (n − r)!
Substitute 6 for n and 2 for r in the given expression, the expression obtained is,
6! / (6 − 2)!
On evaluating the denominator, the expression becomes,
6! / 4!
Now, consider the expression in the numerator, that is 6!
According to the definition of the factorial,
⇒ 6! = 6 × (6 − 1)!
⇒ 6! = 6 × 5!
⇒ 6! = 6 × 5 × 4!
Further, consider the expression
6! / 4!
⇒ 6!/4! = (6 × 5 × 4!)/4!
⇒ 6!/4! = 6 × 5
⇒ 6!/4! = 30
Therefore, the value of the expression n! / (n − r)!, when n = 6, r = 2, is found to be 30.
(ii) n = 9, r = 5
Ans: The expression given is
n! / (n − r)!
Substitute 9 for n and 5 for r in the given expression, the expression obtained is,
9! / (9 − 5)!
On evaluating the denominator, the expression becomes,
9! / 4!
Now, consider the expression in the numerator, that is 9!
According to the definition of the factorial,
⇒ 9! = 9 × (9 − 1)!
⇒ 9! = 9 × 8!
⇒ 9! = 9 × 8 × 7 × 6 × 5 × 4!
Further, consider the expression
9! / 4!
⇒ 9!/4! = (9 × 8 × 7 × 6 × 5 × 4!)/4!
⇒ 9!/4! = 9 × 8 × 7 × 6 × 5
⇒ 9!/4! = 15120
Therefore, the value of the expression n! / (n − r)!, when n = 9, r = 5, is found to be 15120.
FAQs – Class 11 Maths Chapter 6 Exercise 6.2 Permutations and Combinations
Q1. What is the focus of Exercise 6.2?
Exercise 6.2 focuses on permutations, which involve arranging objects where the order is important.
Q2. What is the formula for permutations?
The number of permutations of n distinct objects taken r at a time is:
nPr = n! / (n − r)!
Q3. What is the difference between permutation and combination?
Q4. Why is Exercise 6.2 important for exams?
Permutation-based questions are commonly asked in exams and are important for both board and competitive exams.
Q5. How can students prepare effectively for Exercise 6.2?
Students should practice different types of arrangement problems, understand when to use permutation formulas, and revise factorial concepts for better accuracy.