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Class 11 RD Sharma Solutions- Chapter 16 Permutations – Exercise 16.1

  • Last Updated : 25 Jan, 2021

Question 1. Compute:

(i) 30!/28!

(ii) (11! – 10!)/9!

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

Solution:

(i) 30!/28!

We know that,

n! = n(n-1)!

Therefore, 30! = 30 × 29! = 30 ×29×28!

30!/28! = (30 × 29 × 28!)/28!

= 30 × 29 = 870

(ii) (11! – 10!)/9!

We know that,

n! = n(n-1)!

Therefore,

11! = 11 × 10! =11 × 10 × 9!

10! = 10 × 9!

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!)

We know that,

8! = 8 × 7 × 6!

7! = 7 × 6!

6! = 6!



So,

L.C.M. of 6!, 7!, 8! = LCM [8 × 7 × 6!, 7 × 6!, 6!]

= 8 × 7 × 6!

= 8!

Question 2. Prove that: 1/9! + 1/10! + 1/11! = 122/11!

Solution:

Given:

1/9! + 1/10! + 1/11! = 122/11!

 LHS: 1/9! + 1/10! + 1/11!

Using n = n(n-1)!

1/9! + 1/10! + 1/11! = 1/9! + 1/(10×9!) + 1/(11×10×9!)

= (110 + 11 + 1)/(11 × 10 × 9!)

= 122/11!

= RHS

Therefore, RHS = LHS

Hence, proved.

Question 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!

6! = 6 × 5!



So by using these values

1/4! + 1/5! = x/6!

1/4! + 1/(5×4!) = x/(6×5!)

(5 + 1) / (5×4!) = x/(6×5!)

6/5! = x/(6×5!)

x = (6 × 6 × 5!)/5!

= 36

∴ x = 36.

(ii) x/10! = 1/8! + 1/9!

We know that

10! = 10 × 9!

9! = 9 × 8!

 Using these values, we get

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

∴ x = 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.

Question 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

We can rewrite the above expression 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

We can rewrite the above expression as:

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)

We can rewrite the above expression as:

(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)

We can rewrite the above expression as:

1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1) = [(1) (3) (5) … (2n-1)] [(2) (4) (6) … (2n)] / [(2) (4) (6) … (2n)]

Take 2 out from each term of denominator

= [(1) (2) (3) (4) … (2n-1) (2n)] / 2n [(1) (2) (3) … (n)]

= (2n)! / 2nn!

Question 5. Which of the following are true:

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

(ii) (2 × 3)! = 2! × 3!

Solution:

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

 LHS:

(2 + 3)! = 5!

RHS,

2! + 3! = (2×1) + (3×2×1)

= 2 + 6

= 8

LHS ≠ RHS

∴ The given expression is false.

(ii) (2 × 3)! = 2! × 3!

 LHS:

(2 × 3)! = 6!

= 6 × 5 × 4 × 3 × 2 × 1



= 720

 RHS,

2! × 3! = (2×1) × (3×2×1)

= 12

LHS ≠ RHS

∴ The given expression is false.

Question 6. Prove that: n! (n + 2) = n! + (n + 1)!

Solution:

Given:

n! (n + 2) = n! + (n + 1)!

Method -1 (converting RHS to LHS)

RHS = n! + (n + 1)!

n! + (n + 1)! = n! + (n + 1) (n)!

= n!(1 + n + 1)

= n! (n + 2)

= L.H.S

L.H.S = R.H.S

Hence, Proved.

Method – 2 (converting LHS to RHS)

LHS = n! (n + 2)

n! (n + 2) = n!  (1 + n + 1) 

= n!(1)+n!(n + 1)

= n! +(n + 1)!

= R.H.S

R.H.S = L.H.S 

Hence, Proved.

Question 7. If (n + 2)! = 60[(n – 1)!], find n.

Solution: 

We know that,

n! = n(n-1)!

By using this property

(n + 2)! = 60[(n – 1)!]

(n + 2)(n + 1)(n)[(n – 1)!] = 60[(n – 1)!]

(n + 2)(n + 1)(n) = 60

(n + 2)(n + 1)(n) = 5 × 4 × 3

On comparing both sides, we get

 n = 3

∴ n = 3

Question 8. If (n + 1)! = 90[(n – 1)!], find n.

Solution:

We know that,

n! = n(n-1)!

By using this property

(n + 1)! = 90[(n – 1)!]

(n + 1)(n)[(n – 1)!] = 90[(n – 1)!]

(n + 1)(n) = 90

(n + 1)(n) = 10 × 9

On comparing both sides,we get

n = 9

∴ n = 9

Question 9. If (n + 3)! = 56[(n + 1)!], find n.

Solution:

We know that,

n! = n(n-1)!

By using this property

(n + 3)! = 56[(n + 1)!]

(n + 3)(n+2)[(n + 1)!] = 56[(n + 1)!]

(n + 3)(n + 2) = 56

(n + 3)(n + 2) = 8 × 7

On comparing both sides,we get

n + 2 = 7

∴ n = 5

Question 10. If (2n)! / (3!(2n – 3)!) and n! / (2!(n – 2)!) are in the ratio 44:3, find n.

Solution:

Let [(2n)! / (3!(2n – 3)!)] / [n! / (2!(n – 2)!) ] = 44/3 



[(2n)! × 2!(n – 2)!] / [3!(2n – 3)! × n!] = 44/3

[2n×(2n-1)×(2n-2)×(2n-3)!×2!×(n-2)!] / [3!×(2n-3)!×n×(n-1)×(n-2)!] = 44/3

(2n×(2n-1)×2(n-1))/(3×n×(n-1)) = 44/3

4(2n-1) = 44

2n-1 = 11

2n = 12

∴ n = 6

Question 11. Prove that:

i) n! / (n-r)!  = n(n-1)(n-2) . . . (n-(r-1))

ii) n! / ((n-r)!r!) + n! / ((n-r+1)!(r-1)!) = (n+1)! / (r!(n-r+1)!)

Solution:

(i) n! / (n-r)!  = n(n-1)(n-2) . . . (n-(r-1))

LHS:

n! / (n-r)!  = [n×(n-1)×(n-2) . . . (n-r+2)×(n-r+1)×(n-r)!] / (n-r)!

= n×(n-1)×(n-2). . . (n-r+2)×(n-r+1)

= n×(n-1)×(n-2). . . (n-(r-2))×(n-(r-1))

= n(n-1)(n-2). . . (n-(r-1))

= RHS

LHS = RHS

Hence, proved.

(ii) n! / ((n-r)!r!) + n! / ((n-r+1)!(r-1)!) = (n+1)! / (r!(n-r+1)!)

LHS:

n! / ((n-r)!r!) + n! / ((n-r+1)!(r-1)!) =  n![(n-r+1) / ((n-r+1)!r!) + r / ((n-r+1)!r!) ]

= n![(n-r+1+r) / (n-r+1)!r!]

= n!(n+!)/[(n-r+1)!r!]

= (n+1)! / (r!(n-r+1)!)

= RHS

LHS = RHS

Hence, proved.

Question 12. Prove that: i) (2n+1)! / n!  = 2n [1.3.5. . . (2n – 1)(2n + 1)]

Solution:

LHS:

(2n+1)! / n! = [1×2×3×4 . . . (2n-2)×(2n-1)×(2n)×(2n+1)] / n!

Separate even and odd terms in numerator

= {[2×4 ×6 . . . (2n-2)×(2n)][1×3 ×5 . . . (2n-1)×(2n+1)]} / n!

Take factor of two out from all even terms of numerator

= {2n [1×2×3. . . (n-1)×(n)] [1×3×5. . . (2n-1)×(2n+1)]} / n!

= {2n n! [1×3 ×5. . . (2n-1)×(2n+1)]}/n!

= 2n [1×3×5. . . (2n-1)×(2n+1)]

= RHS

LHS = RHS

Hence, proved.                        


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