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NCERT Solutions For Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.2

Here, Below you all know about NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.2 Question Answer. I know many of you confuse about finding Chapter 7 Permutations and Combinations Ex 7.2 Of Class 11 NCERT Solutions. So, Read the full post below and get your solutions.

TextbookNCERT
BoardCBSE
CategoryNCERT Solutions
ClassClass 11
SubjectMaths
ChapterChapter 7
ExerciseClass 11 Maths Chapter 7 Permutations and Combinations Exercise 7.2
Number of Questions Solved5

NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.2

NCERT TEXTBOOK EXERCISES

Ex 7.2 Class 11 Maths Question 1.
Evaluate
(i) 8!
(ii) 4!-3!

Solution.

(i) 8! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40320
(ii) 4! – 3! = (1 x 2 x 3 x 4) – (1 x 2 x 3) = 24 – 6 = 18

Ex 7.2 Class 11 Maths Question 2.
Is 3! + 4! = 7! ?

Solution.

3! + 4! = (1 x 2 x 3) + (1 x 2 x 3 x 4) = 6 + 24 = 30 … (i)
7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040 ………(ii)
From (i) & (ii), we get 3! + 4! ≠ 7!.

Ex 7.2 Class 11 Maths Question 3.
$\frac { 8! }{ 6!\times 2! }$

Solution.

$\frac { 8! }{ 6!\times 2! } =\frac { 8\times 7\times 6! }{ 6!\times 2! } =4\times 7=28$

Ex 7.2 Class 11 Maths Question 4.
$\frac { 1 }{ 6! } +\frac { 1 }{ 7! } =\frac { x }{ 8! }$, find.

Solution.

We have, $\frac { 1 }{ 6! } +\frac { 1 }{ 7! } =\frac { x }{ 8! }$

Ex 7.2 Class 11 Maths Question 5.
Evaluate $\frac { n! }{ \left( n-r \right) ! }$, when
(i) n = 6, r = 2
(ii) n = 9, r = 5

Solution.

(i) $\frac { 6! }{ \left( 6-2 \right) ! } =\frac { 6! }{ 4! } =\frac { 6\times 5\times 4! }{ 4! } =30$

(ii) $\frac { 9! }{ \left( 9-5 \right) ! } =\frac { 9! }{ 4! } =\frac { 9\times 8\times 7\times 6\times 5\times 4! }{ 4! } =15120$

NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Exercise 7.2 PDF

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