RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.1: In this exercise, we will introduce the term and notation for a factorial. Our solution module uses several tips and shortcut diagrams to explain all exercise problems in the most understandable language. Subject matter experts have simplified difficult problems into simple steps, which students can easily solve with accuracy. RD Sharma Class 11 Mathematics Chapter 16 Solutions for the chapter Permutations will help students gain solid knowledge and mastery of the subject. Students can easily download the PDF from the link provided below.
Download RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.1 PDF:
Download Class 11 RD Sharma Solutions Maths Chapter 16 Exercise 16.1
Access RD Sharma Solutions Class 11 Maths Chapter 16
1. Compute:
(i) 30!/28!
(ii) (11! – 10!)/9!
(iii) L.C.M. (6!, 7!, 8!)
Solution:
(i) 30!/28!
Let us evaluate,
30!/28! = (30 × 29 × 28!)/28!
= 30 × 29
= 870
(ii) (11! – 10!)/9!
Let us evaluate,
We know,
11! = 11 × 10 × 9 × …. × 1
10! = 10 × 9 × 8 × … × 1
9! = 9 × 8 × 7 × … × 1
By using these values we get,
(11! – 10!)/9! = (11 × 10 × 9! – 10 × 9!)/ 9!
= 9! (110 – 10)/9!
= 110 – 10
= 100
(iii) L.C.M. (6!, 7!, 8!)
Let us find the LCM of (6!, 7!, 8!)
We know,
8! = 8 × 7 × 6!
7! = 7 × 6!
6! = 6!
So,
L.C.M. (6!, 7!, 8!) = LCM [8 × 7 × 6!, 7 × 6!, 6!]
= 8 × 7 × 6!
= 8!
2. Prove that: 1/9! + 1/10! + 1/11! = 122/11!
Solution:
Given:
1/9! + 1/10! + 1/11! = 122/11!
Let us consider LHS: 1/9! + 1/10! + 1/11!
1/9! + 1/10! + 1/11! = 1/9! + 1/(10×9!) + 1/(11×10×9!)
= (110 + 11 + 1)/(11 × 10 × 9!)
= 122/11!
= RHS
Hence proved.
3. Find x in each of the following:
(i) 1/4! + 1/5! = x/6!
(ii) x/10! = 1/8! + 1/9!
(iii) 1/6! + 1/7! = x/8!
Solution:
(i) 1/4! + 1/5! = x/6!
We know that
5! = 5 × 4 × 3 × 2 × 1
6! = 6 × 5 × 4 × 3 × 2 × 1
So by using these values,
1/4! + 1/5! = x/6!
1/4! + 1/(5×4!) = x/6!
(5 + 1) / (5×4!) = x/6!
6/5! = x/(6×5!)
x = (6 × 6 × 5!)/5!
= 36
∴ The value of x is 36.
(ii) x/10! = 1/8! + 1/9!
We know that
10! = 10 × 9!
9! = 9 × 8!
So by using these values,
x/10! = 1/8! + 1/9!
x/10! = 1/8! + 1/(9×8!)
x/10! = (9 + 1) / (9×8!)
x/10! = 10/9!
x/(10×9!) = 10/9!
x = (10 × 10 × 9!)/9!
= 10 × 10
= 100
∴ The value of x is 100.
(iii) 1/6! + 1/7! = x/8!
We know that
8! = 8 × 7 × 6!
7! = 7 × 6!
So by using these values,
1/6! + 1/7! = x/8!
1/6! + 1/(7×6!) = x/8!
(1 + 7)/(7×6!) = x/8!
8/7! = x/8!
8/7! = x/(8×7!)
x = (8 × 8 × 7!)/7!
= 8 × 8
= 64
∴ The value of x is 64.
4. Convert the following products into factorials:
(i) 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10
(ii) 3 ⋅ 6 ⋅ 9 ⋅ 12 ⋅ 15 ⋅ 18
(iii) (n + 1) (n + 2) (n + 3) …(2n)
(iv) 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1)
Solution:
(i) 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10
Let us evaluate
We can write it as:
5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10 = (1×2×3×4×5×6×7×8×9×10)/(1×2×3×4)
= 10!/4!
(ii) 3 ⋅ 6 ⋅ 9 ⋅ 12 ⋅ 15 ⋅ 18
Let us evaluate
3 ⋅ 6 ⋅ 9 ⋅ 12 ⋅ 15 ⋅ 18 = (3×1) × (3×2) × (3×3) × (3×4) × (3×5) × (3×6)
= 36 (1×2×3×4×5×6)
= 36 (6!)
(iii) (n + 1) (n + 2) (n + 3) … (2n)
Let us evaluate
(n + 1) (n + 2) (n + 3) … (2n) = [(1) (2) (3) .. (n) … (n + 1) (n + 2) (n + 3) … (2n)] / (1) (2) (3) .. (n)
= (2n)!/n!
(iv) 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1)
Let us evaluate
1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1) = [(1) (3) (5) … (2n-1)] [(2) (4) (6) … (2n)] / [(2) (4) (6) … (2n)]
= [(1) (2) (3) (4) … (2n-1) (2n)] / 2n [(1) (2) (3) … (n)]
= (2n)! / 2n n!
5. Which of the following is true:
(i) (2 + 3)! = 2! + 3!
(ii) (2 × 3)! = 2! × 3!
Solution:
(i) (2 + 3)! = 2! + 3!
Let us consider LHS: (2 + 3)!
(2 + 3)! = 5!
Now RHS,
2! + 3! = (2×1) + (3×2×1)
= 2 + 6
= 8
LHS ≠ RHS
∴ The given expression is false.
(ii) (2 × 3)! = 2! × 3!
Let us consider LHS: (2 × 3)!
(2 × 3)! = 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720
Now RHS,
2! × 3! = (2×1) × (3×2×1)
= 12
LHS ≠ RHS
∴ The given expression is false.
6. Prove that: n! (n + 2) = n! + (n + 1)!
Solution:
Given:
n! (n + 2) = n! + (n + 1)!
Let us consider RHS = n! + (n + 1)!
n! + (n + 1)! = n! + (n + 1) (n + 1 – 1)!
= n! + (n + 1)n!
= n!(1 + n + 1)
= n! (n + 2)
= L.H.S
L.H.S = R.H.S
Hence, Proved.
We have included all the information regarding CBSE RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.1. If you have any queries feel free to ask in the comment section.
FAQ: RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.1
Can I open RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.1 PDF on my smartphone?
Yes, you can open RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.1 PDF on any device.
What are the benefits of studying from RD Sharma Solutions Class 12?
By practicing these solutions, students can earn higher academic grades. Our experts solve these solutions with utmost accuracy to help students in their studies.
Can I download RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.1 PDF free?
Yes, you can download RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.1 PDF free.
Is RD Sharma enough for Class 12 Maths?
RD Sharma is a good book that gives you thousands of questions to practice.