FUZY MATH ACADEMY

Class 11 Maths Chapter 6 Permutations and Combinations – Exercise 6.2 NCERT Solutions Introduction Exercise 6.2 introduces the concept of combinations. Unlike permutations, combinations focus on selections where order does not matter. You will learn how to calculate the number of ways to choose objects from a set and apply these ideas to practical problems. Key Concepts Combination Formula: Number of combinations of n objects taken r at a time: C ( n , r ) = ( n r ) = n ! r ! ( n − r ) ! Properties of Combinations: ( n 0 ) = 1 ( n n ) = 1 ( n r ) = ( n n − r ) Difference from Permutations: Permutations: order matters. Combinations: order does not matter. NCERT Questions with Solutions (10) Q1. Find number of combinations of 5 objects taken 3 at a time. C ( 5 , 3 ) = 5 ! 3 ! ⋅ 2 ! = 120 12 = 10 Q2. Find number of combinations of 6 objects taken 2 at a time. C ( 6 , 2 ) = 6 ! 2 ! ⋅ 4 ! = 720 48 = 15 Q3. Find number of combinations of 4 objects taken all at a time. C ( 4 , 4 ) = 1 Q4....

Class 11 Maths Chapter 6 Permutations and Combinations – Exercise 6.1 NCERT Solutions Introduction Exercise 6.1 introduces the concept of permutations. You will learn how to count arrangements of objects, apply factorial notation, and solve problems involving ordered arrangements. This exercise builds the foundation for combinatorics and probability. Key Concepts Factorial Notation: n ! = n ⋅ ( n − 1 ) ⋅ ( n − 2 ) ⋯ 3 ⋅ 2 ⋅ 1 Permutation Formula: Number of permutations of n objects taken r at a time: P ( n , r ) = n ! ( n − r ) ! Special Cases: P ( n , n ) = n ! P ( n , 1 ) = n NCERT Questions with Solutions (10) Q1. Find number of permutations of 5 objects taken 3 at a time. P ( 5 , 3 ) = 5 ! ( 5 − 3 ) ! = 120 2 = 60 Q2. Find number of permutations of 6 objects taken 2 at a time. P ( 6 , 2 ) = 6 ! ( 6 − 2 ) ! = 720 24 = 30 Q3. Find number of permutations of 4 objects taken all at a time. P ( 4 , 4 ) = 4 ! = 24 Q4. Find number of permutations of 7 objects taken 1 at a time. P ( 7 ...

Class 11 Maths Chapter 7 Binomial Theorem – Exercise 7.2 NCERT Solutions Introduction Exercise 7.2 focuses on identifying specific terms in binomial expansions. You will learn how to find the general term, middle term(s), and coefficients of required terms in expansions using the Binomial Theorem. This exercise strengthens your ability to apply the theorem in targeted problems. Key Concepts General Term in Expansion: T r + 1 = ( n r ) x n − r y r Middle Term(s): If n is even: single middle term at r = n 2 . If n is odd: two middle terms at r = n − 1 2 , n + 1 2 . Coefficient of Term: Coefficient of x n − r y r is ( n r ) . NCERT Questions with Solutions (10) Q1. Find general term in expansion of ( x + y ) n . T r + 1 = ( n r ) x n − r y r Q2. Find 5th term in expansion of ( x + y ) 7 . T 5 = ( 7 4 ) x 3 y 4 = 35 x 3 y 4 Q3. Find 4th term in expansion of ( a + b ) 6 . T 4 = ( 6 3 ) a 3 b 3 = 20 a 3 b 3 Q4. Find middle term in expansion of ( x + y ) 8 . Since n = 8 (even), midd...

Class 11 Maths Chapter 7 Binomial Theorem – Exercise 7.1 NCERT Solutions Introduction Exercise 7.1 introduces the Binomial Theorem, which provides a formula for expanding expressions of the form ( x + y ) n . You will learn how to apply the theorem, calculate coefficients, and solve problems involving binomial expansions. This exercise is fundamental for algebra and probability. Key Concepts Binomial Theorem: For any positive integer n : ( x + y ) n = ∑ r = 0 n ( n r ) x n − r y r Binomial Coefficient: ( n r ) = n ! r ! ( n − r ) ! Properties: Total number of terms = n + 1 . First term = x n . Last term = y n . Middle term(s) depend on whether n is even or odd. NCERT Questions with Solutions (10) Q1. Expand ( x + y ) 5 . ( x + y ) 5 = x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 x y 4 + y 5 Q2. Expand ( a + b ) 4 . ( a + b ) 4 = a 4 + 4 a 3 b + 6 a 2 b 2 + 4 a b 3 + b 4 Q3. Expand ( p + q ) 3 . ( p + q ) 3 = p 3 + 3 p 2 q + 3 p q 2 + q 3 Q4. Expand ( x − 1 ) 4 . ( x − 1 ) 4 = x 4 − 4...

Class 11 Maths Chapter 8 Sequences and Series – Exercise 8.4 NCERT Solutions Introduction Exercise 8.4 focuses on the sum to n terms of special series. Students learn how to evaluate sums of series like ∑ n , ∑ n 2 , and ∑ n 3 , and apply these formulas to solve problems. This exercise is essential for building problem‑solving skills in sequences and series. Key Formulas Sum of first n natural numbers: 1 + 2 + 3 + ⋯ + n = n ( n + 1 ) 2 Sum of squares of first n natural numbers: 1 2 + 2 2 + 3 2 + ⋯ + n 2 = n ( n + 1 ) ( 2 n + 1 ) 6 Sum of cubes of first n natural numbers: 1 3 + 2 3 + 3 3 + ⋯ + n 3 = ( n ( n + 1 ) 2 ) 2 Common Mistakes Forgetting to apply the correct formula for squares or cubes. Mixing up ∑ n 2 and ∑ n 3 . Not simplifying expressions properly. Ignoring factorization in final answers. NCERT Questions with Step‑by‑Step Solutions (10) Q1. Find sum of first 10 natural numbers. S = 10 ( 10 + 1 ) 2 = 55 Q2. Find sum of squares of first 7 natural numbers. S = 7 ( 7 + 1 ) (...

Class 11 Maths Chapter 8 Sequences and Series – Exercise 8.3 NCERT Solutions Introduction Exercise 8.3 introduces the concept of geometric progression (GP). You will learn how to identify GPs, find common ratios, calculate general terms, and evaluate sums of finite and infinite geometric series. This exercise builds the foundation for advanced series and summation techniques in mathematics. Key Concepts Geometric Progression (GP): A sequence where each term after the first is obtained by multiplying with a constant ratio. a , a r , a r 2 , a r 3 , … where a = first term, r = common ratio. General Term of GP: a n = a ⋅ r n − 1 Sum of First n Terms of GP: S n = a ( r n − 1 ) r − 1 , r ≠ 1 Sum of Infinite GP: If ∣ r ∣ < 1 : S ∞ = a 1 − r NCERT Questions with Solutions (10) Q1. Find 7th term of GP: 2, 6, 18, … a = 2 ,   r = 3 ,   a 7 = 2 ⋅ 3 6 = 1458 Q2. Find 10th term of GP: 5, 10, 20, … a = 5 ,   r = 2 ,   a 10 = 5 ⋅ 2 9 = 2560 Q3. Find 12th term of GP: 3, 9, 27, … a = 3 ,   r...

Class 11 Maths Chapter 8 Sequences and Series – Exercise 8.2 NCERT Solutions Introduction Exercise 8.2 focuses on the sum of terms in an arithmetic progression (AP). You will learn how to derive the formula for the sum of first n terms, apply it to practical problems, and solve questions involving partial sums. This exercise builds the foundation for advanced series and summation techniques. Key Concepts Sum of First n Terms of AP: S n = n 2 [ 2 a + ( n − 1 ) d ] or equivalently, S n = n 2 ( a + l ) where a = first term, d = common difference, l = last term. Special Cases: If d = 0 , sum = n ⋅ a . If d > 0 , AP is increasing; if d < 0 , AP is decreasing. NCERT Questions with Solutions (10) Q1. Find sum of first 10 terms of AP: 2, 7, 12, … a = 2 ,   d = 5 ,   n = 10 S 10 = 10 2 [ 2 ⋅ 2 + ( 10 − 1 ) ⋅ 5 ] = 5 [ 4 + 45 ] = 245 Q2. Find sum of first 20 terms of AP: 1, 4, 7, … a = 1 ,   d = 3 ,   n = 20 S 20 = 20 2 [ 2 ⋅ 1 + ( 20 − 1 ) ⋅ 3 ] = 10 [ 2 + 57 ] = 590 Q3. Find sum of ...