NCERT Solutions Class 11 Mathematics Permutations Combinations Exercise 7.2

Class 11 - Mathematics
Permutations Combinations - Exercise 7.2

NCERT Solutions class 11 Mathematics Textbook
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Question 1:
Evaluate (i) 8!                                         (ii) 4! – 3!

Answer:
We know that
n! = n(n – 1)(n – 2)(n – 3)……………..3.2.1
or
n! = 1.2.3………..(n – 2)(n – 1)n
(i) 8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40320
(ii) 4! = 4 * 3 * 2 * 1 = 24
     3! = 3 * 2 * 1 = 6
Now, 4! – 3! = 24 – 6 = 18


Question 2:
Is 3! + 4! = 7!?

Answer:
We know that
n! = n(n – 1)(n – 2)(n – 3)……………..3.2.1
or
n! = 1.2.3………..(n – 2)(n – 1)n
3! = 1 * 2 * 3 = 6
4! = 1 * 2 * 3 * 4 = 24
Now, 3! + 4! = 6 + 24 = 30
7! = 1 * 2 * 3 * 4 * 5 * 6 * 7 = 5040     So, 3! + 4! ≠ 7!


Question 3:
Compute 8!/(6! * 2!)

Answer:
8!/(6! * 2!) = (8 * 7 * 6!)/(6! * 2 * 1) = 56/2 = 28

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Question 4:
If 1/6! + 1/7! = x/8!, find x.

Answer:
Given, 1/6! + 1/7! = x/8!
=> 1/6! + 1/(7 * 6!) = x/(8 * 7 * 6!)
=> (1/6!)(1 + 1/7) = x/(8 * 7 * 6!)
=> 1 + 1/7 = x/56
=> 8/7 = x/56
=> x = 8 * 56
=> x = (8 * 56)/8
=> x = 8 * 8
=> x = 64
So, the value of x is 64.


Question 5:
Evaluate n!/(n – r)!, when
(i) n = 6, r = 2                                  (ii) n = 9, r = 5

Answer:
 (i) Given, n = 6, r = 2
Now, n!/(n – r)! = 6!/(6 – 2)! = 6!/4! = (6 * 5 * 4!)/4! = 6 * 5 = 30                                 
(ii) Given, n = 9, r = 5
Now, n!/(n – r)! = 9!/(9 – 5)! = 9!/4!
                                                  = (9 * 8 * 7 * 6 * 5 * 4!)/4!
                                                  = 9 * 8 * 7 * 6 * 5
                                                  = 15120

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