Class 11 Maths Ncert Solutions Ex 7.2

Class 11 Maths Ncert Solutions Chapter 7 Ex 7.2

Q-1: Evaluate:

(i). 8!                                                                    (ii). 4! – 3!

 

Solution:

(i). 8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320

(ii). 4! = 1 × 2 × 3 × 4 = 24

And, 3! = 1 × 2 × 3 = 6

Therefore, 4! – 3! = 24 – 6 = 18

Q-2: Is 3! + 4! = 7! ?

 

Solution:

Now, 3! = 1 × 2 × 3 = 6

4! = 1 × 2 × 3 × 4 = 24

And, 7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040

Therefore, 4! – 3! ≠ 7!

Q-3: Compute:

9!5!×3!

Solution:

9!5!×3! = 1×2×3×4×5×6×7×8×91×2×3×4×5×1×2×3

Therefore, 9!5!×3! = 392

 

 

Q-4. If 15!+16!=a7!, find the value of a.

 

Solution:

15!+16!=a7!

   15!+16×5!=a7×6×5!

   15!+[1+16]=a7×6×5!

1+16=a7×6

   76=a7×6

a=7×7×66

Therefore, a = 49

Q-5: Evaluate: a!(ar)!, when

(i). a = 5, r = 1                                                       (ii). a = 8, r = 4

 

Solution:

(i)  a = 5, r = 1:         [ Given ]

a!(ar)! = 5!(51)!= 5!4!= 5×4!4!

Therefore, a!(ar)! = 5

(ii) a!(ar)! = 8!(84)!= 8!4!

= 8×7×6×5×4!4!= 8 × 7 × 6 × 5 = 15120

Therefore, a!(ar)! = 15120