Class 11 NCERT Solutions Chapter 7 Permutations And Combinations Exercise 7.2 (original) (raw)
Last Updated : 23 Jul, 2025
Chapter 7 of the Class 11 NCERT Mathematics textbook, titled "Permutations and Combinations," delves into the principles of counting and arranging objects. This chapter covers the concepts of permutations and combinations, which are crucial for solving problems involving the arrangement and selection of objects. Exercise 7.2 presents problems that require applying these concepts to determine the number of permutations and combinations in various scenarios.
NCERT Solutions for Class 11 - Chapter 7 Permutations and Combinations - Exercise 7.2
This section provides detailed solutions for Exercise 7.2 from Chapter 7 of the Class 11 NCERT Mathematics textbook. The exercise includes problems that involve calculating permutations and combinations with more complex conditions, such as repeated elements and restrictions. Solutions are presented step-by-step to help students understand the methods for solving these problems and applying permutation and combination principles effectively.
**Question 1. Evaluate.
**i) 8!
**Solution:
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 56 × 30 × 12 × 2
= 1680 × 24
= 40320
**ii) 4! – 3!
**Solution:
4! -3! = 4 × 3! - 3!
= 3! × (4 - 1)
= 3! × 3
= 6 × 3
= 18
**Question 2. Is 3! + 4! = 7! ?
**Solution:
L.H.S = 3! + 4!
= 3! + 4 * 3!
= 3! * (1 + 4)
= 3! * 5
= 60
R.H.S = 7!
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 42 × 20 × 6
= 42 × 120
= 5040
As l.H.S ≠ R.H.S
**NO, 3!+ 4! is not equal to 7!
**Question 3. Compute 8!/(6! × 2!)
**Solution:
8!/(6! × 2!)
= 8 × 7 × 6 ! / (6 ! × 2!)
Now, both 6! on numerator and denominator will be cancelled
= 8 × 7/2 × 1
= 4 × 7
= 28
**Question 4. If (1/6!) + (1/7!) = (y/8!), Find y.
**Solution:
L.C.M of 6! and 7! is 7!
We can write the equation like this,
(7 + 1)/7! = (y/8!)
8 × 8!/7 ! = y
8 × 8 × 7! / 7! = y
8 × 8 = y
y = 64
Hence, value of y is 64
**Question 5. Evaluate n!/(n - r)! when
**i) n = 6, r = 2
**Solution:
n!/(n - r)!
Here n = 6 and r = 2, we have
6!/(6 - 2)!
= 6!/4!
= 6 × 5 × 4!/4!
= 6 × 5
= 30
**ii) n = 9, r = 5
**Solution:
n!/(n - r)!
Here n = 9 and r = 5, we have
9!/(9 - 5)!
= 9!/4!
= 9 × 8 × 7 × 6 × 5 × 4!/4!
= 9 × 8 ×7 × 6 × 5
= 72 × 7 × 30
= 72 × 210
= 15120
Summary
Chapter 7 of the Class 11 NCERT Mathematics textbook, "Permutations and Combinations," explores the principles of counting and arranging objects. Exercise 7.2 includes problems that involve calculating permutations and combinations with complex conditions, such as repetitions and restrictions. The exercise helps students apply permutation and combination concepts to solve advanced problems, reinforcing their understanding through step-by-step solutions.