**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)

= 3^{6} (1×2×3×4×5×6)

= 3^{6} (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)] / 2^{n} [(1) (2) (3) … (n)]

= (2n)! / 2^{n} 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.

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