Choice (a) is correct. We have, (a³ -3a²b + 3ab² - b³)⁵⁰ = {(a - b)³}⁵⁰ = (a - b)¹⁵⁰ We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1). Total number of terms in the expansion of (a - b)¹⁵⁰ is (150 + 1) = 151. Total number of terms in the expansion of (a³ - 3a²b + 3ab² -Read more
Choice (a) is correct.
We have, (a³ -3a²b + 3ab² – b³)⁵⁰ = {(a – b)³}⁵⁰ = (a – b)¹⁵⁰
We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1).
Total number of terms in the expansion of (a – b)¹⁵⁰ is (150 + 1) = 151.
Total number of terms in the expansion of (a³ – 3a²b + 3ab² – b³)⁵⁰ is 151.
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
Choice (c) is correct We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1). Total number of terms in the expansion of (x + a)⁵⁰¹ is (501 + 1) = 502. Total number of terms in the expansion of (x + a)⁵⁰¹ + (x - a)⁵⁰¹ is 251 terms get cancelled. This question related to ChapteRead more
Choice (c) is correct
We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1).
Total number of terms in the expansion of (x + a)⁵⁰¹ is (501 + 1) = 502.
Total number of terms in the expansion of (x + a)⁵⁰¹ + (x – a)⁵⁰¹ is 251 terms get cancelled.
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
Choice (b) is correct. We have, (a² + 2ab + b²)¹⁰¹ = {(a + b)² }¹⁰¹ = (a + b) ²⁰² We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1). Total number of terms in the expansion of (a + b) ²⁰² is (202 + 1) = 203. Total number of terms in the expansion of (a² + 2ab + b²) ¹⁰¹ iRead more
Choice (b) is correct.
We have, (a² + 2ab + b²)¹⁰¹ = {(a + b)² }¹⁰¹ = (a + b) ²⁰²
We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1).
Total number of terms in the expansion of (a + b) ²⁰² is (202 + 1) = 203.
Total number of terms in the expansion of (a² + 2ab + b²) ¹⁰¹ is 203.
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
Choice (b) is correct. 3 different rings to be worn in four fingers with at most one in each finger in P(4, 3) ways = 4!/(4 - 3)! = 4!/1! = 4 × 3 × 2 × 1/1 = 24 This question related to Chapter 5 maths Class 11th NCERT. From the Chapter 6: Permutations and Combinations. Give answer according to yourRead more
Choice (b) is correct.
3 different rings to be worn in four fingers with at most one in each finger in P(4, 3) ways = 4!/(4 – 3)! = 4!/1! = 4 × 3 × 2 × 1/1 = 24
This question related to Chapter 5 maths Class 11th NCERT. From the Chapter 6: Permutations and Combinations. Give answer according to your understanding.
Choice (c) is correct. Given ¹⁵ C₃ᵣ = ¹⁵ Cᵣ ₊ ₇ ⇒ 3r = r + 7 or 3r + ( r + 7) = 15 ⇒ 2r = 7 or 4r = 8 ⇒ r = 7/2 or r = 2 ⇒ r = 2 This question related to Chapter 6 maths Class 11th NCERT. From the Chapter 6: Permutations and Combinations. Give answer according to your understanding. For more pleasRead more
Choice (c) is correct.
Given ¹⁵ C₃ᵣ = ¹⁵ Cᵣ ₊ ₇
⇒ 3r = r + 7 or 3r + ( r + 7) = 15
⇒ 2r = 7 or 4r = 8
⇒ r = 7/2 or r = 2
⇒ r = 2
This question related to Chapter 6 maths Class 11th NCERT. From the Chapter 6: Permutations and Combinations. Give answer according to your understanding.
The total number of terms in the expansion of (a³ -3a²b + 3ab² – b³)⁵⁰ is:
Choice (a) is correct. We have, (a³ -3a²b + 3ab² - b³)⁵⁰ = {(a - b)³}⁵⁰ = (a - b)¹⁵⁰ We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1). Total number of terms in the expansion of (a - b)¹⁵⁰ is (150 + 1) = 151. Total number of terms in the expansion of (a³ - 3a²b + 3ab² -Read more
Choice (a) is correct.
We have, (a³ -3a²b + 3ab² – b³)⁵⁰ = {(a – b)³}⁵⁰ = (a – b)¹⁵⁰
We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1).
Total number of terms in the expansion of (a – b)¹⁵⁰ is (150 + 1) = 151.
Total number of terms in the expansion of (a³ – 3a²b + 3ab² – b³)⁵⁰ is 151.
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-7
The total number of terms in the expansion of (x + a)⁵⁰¹ + (x – a)⁵⁰¹ is :
Choice (c) is correct We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1). Total number of terms in the expansion of (x + a)⁵⁰¹ is (501 + 1) = 502. Total number of terms in the expansion of (x + a)⁵⁰¹ + (x - a)⁵⁰¹ is 251 terms get cancelled. This question related to ChapteRead more
Choice (c) is correct
We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1).
Total number of terms in the expansion of (x + a)⁵⁰¹ is (501 + 1) = 502.
Total number of terms in the expansion of (x + a)⁵⁰¹ + (x – a)⁵⁰¹ is 251 terms get cancelled.
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-7
The total number of terms in the expansion of (a² + 2ab + b²)¹⁰¹ is :
Choice (b) is correct. We have, (a² + 2ab + b²)¹⁰¹ = {(a + b)² }¹⁰¹ = (a + b) ²⁰² We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1). Total number of terms in the expansion of (a + b) ²⁰² is (202 + 1) = 203. Total number of terms in the expansion of (a² + 2ab + b²) ¹⁰¹ iRead more
Choice (b) is correct.
We have, (a² + 2ab + b²)¹⁰¹ = {(a + b)² }¹⁰¹ = (a + b) ²⁰²
We know that the total number of terms in the expansion of (x + a)ⁿ is (n + 1).
Total number of terms in the expansion of (a + b) ²⁰² is (202 + 1) = 203.
Total number of terms in the expansion of (a² + 2ab + b²) ¹⁰¹ is 203.
This question related to Chapter 7 maths Class 11th NCERT. From the Chapter 7: Binomial Theorem. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-7
There are 3 different rings to be worn in four fingers with at most one in each finger. Find the number of ways in which it can be done.
Choice (b) is correct. 3 different rings to be worn in four fingers with at most one in each finger in P(4, 3) ways = 4!/(4 - 3)! = 4!/1! = 4 × 3 × 2 × 1/1 = 24 This question related to Chapter 5 maths Class 11th NCERT. From the Chapter 6: Permutations and Combinations. Give answer according to yourRead more
Choice (b) is correct.
3 different rings to be worn in four fingers with at most one in each finger in P(4, 3) ways = 4!/(4 – 3)! = 4!/1! = 4 × 3 × 2 × 1/1 = 24
This question related to Chapter 5 maths Class 11th NCERT. From the Chapter 6: Permutations and Combinations. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-6
If ¹⁵ C₃ᵣ = ¹⁵ Cᵣ ₊ ₇, then the value of r is
Choice (c) is correct. Given ¹⁵ C₃ᵣ = ¹⁵ Cᵣ ₊ ₇ ⇒ 3r = r + 7 or 3r + ( r + 7) = 15 ⇒ 2r = 7 or 4r = 8 ⇒ r = 7/2 or r = 2 ⇒ r = 2 This question related to Chapter 6 maths Class 11th NCERT. From the Chapter 6: Permutations and Combinations. Give answer according to your understanding. For more pleasRead more
Choice (c) is correct.
Given ¹⁵ C₃ᵣ = ¹⁵ Cᵣ ₊ ₇
⇒ 3r = r + 7 or 3r + ( r + 7) = 15
⇒ 2r = 7 or 4r = 8
⇒ r = 7/2 or r = 2
⇒ r = 2
This question related to Chapter 6 maths Class 11th NCERT. From the Chapter 6: Permutations and Combinations. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-6