NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.1

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Class 11 Maths NCERT Solutions Chapter 7 Permutations and Combinations Exercise 7.1 is based on following concepts:

Introduction to Permutations and Combinations
Fundamental Principle of Counting

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Check out the solutions of Class 11 Maths NCERT solutions Chapter 7 Permutations and Combinations Exercise 7.1

Read More: NCERT Solutions For Class 11 Maths Chapter 7 Permutations and Combinations

Also check other Exercise Solutions of Class 11 Maths Chapter 7 Permutations and Combinations

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Chapter Related Topics
Permutations and Combinations Formulas	Factorial Formula	Stirling Formula
Combinatorics	Involute	Difference between Permutation and Combination

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CBSE CLASS XII Related Questions

1.
For a square matrix \(A\), \[ (3A)^{-1}= \]
- \( 3A^{-1} \)
- \( 9A^{-1} \)
- \( \frac{1}{3} A^{-1} \)
- \( \frac{1}{9} A^{-1} \)
View Solution
2.
Check whether the function \[ f(x)= \begin{cases} \dfrac{|x-3|}{2(x-3)}, & x<3,\\[6pt] \dfrac{x-6}{6}, & x\geq 3 \end{cases} \] is continuous at \(x=3\) or not.
View Solution
3.
For two vectors \(\vec{a}\) and \(\vec{b}\):

Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]
- Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution
4.
The domain of \[ f(x)=\cos^{-1}(2x-5) \] is:
- \([-1, 1]\)
- \([4, 6]\)
- \([-7, -3]\)
- \([2, 3]\)
View Solution
5.
Check whether \[ f:\mathbb{R}-\{3\}\rightarrow\mathbb{R} \] defined as \[ f(x)=\frac{x-2}{x-3} \] is onto or not.
View Solution
6.
If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is:
- \(\frac{1}{3}\)
- \(\frac{1}{9}\)
- \(3\)
- \(9\)
View Solution

Similar Mathematics Concepts

Permutations and Combinations: Formulas & Examples

Factorial: Definition, Formula, Factorial Table & Examples

Permutations and Combinations Formulas: Derivation, Example

NCERT Solutions for Class 11 Maths Chapter 7: Permutations and Combinations

Combinatorics: Formula, Features, Permutation & Combination

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Involute: Introduction, Equations & Sample Questions

Important Questions For Class 11 Maths Chapter 7: Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.2

NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.3

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