RD Chapter 16- Permutations Ex-16.2 |
RD Chapter 16- Permutations Ex-16.3 |
RD Chapter 16- Permutations Ex-16.4 |
RD Chapter 16- Permutations Ex-16.5 |

Compute:

(i) 30!/28!

(ii) (11! – 10!)/9!

(iii) L.C.M. (6!, 7!, 8!)

**Answer
1** :

(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!

**Answer
2** :

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.

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!

**Answer
3** :

(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.

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)

**Answer
4** :

(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!

Which of the following are true:

(i) (2 + 3)! = 2! + 3!

(ii) (2 × 3)! = 2! × 3!

**Answer
5** :

(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.

**Answer
6** :

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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