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

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

Solution:

(i) 30!/28!

We know that,

n! = n(n-1)!

Therefore, 30! = 30 Ã— 29! = 30 Ã—29Ã—28!

30!/28! = (30 Ã— 29 Ã— 28!)/28!

= 30 Ã— 29 = 870

(ii) (11! â€“ 10!)/9!

We know that,

n! = n(n-1)!

Therefore,

11! = 11 Ã— 10! =11 Ã— 10 Ã— 9!

10! = 10 Ã— 9!

By using these values, we get

(11! â€“ 10!)/9! = (11 Ã— 10 Ã— 9! â€“ 10 Ã— 9!)/ 9!

= 9! (110 â€“ 10)/9!

= 110 â€“ 10

= 100

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

We know that,

8! = 8 Ã— 7 Ã— 6!

7! = 7 Ã— 6!

6! = 6!

So,

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

= 8 Ã— 7 Ã— 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/(10Ã—9!) + 1/(11Ã—10Ã—9!)

= (110 + 11 + 1)/(11 Ã— 10 Ã— 9!)

= 122/11!

= RHS

Therefore, RHS = LHS

Hence, proved.

### (iii) 1/6! + 1/7! = x/8!

Solution:

(i) 1/4! + 1/5! = x/6!

We know that

5! = 5 Ã— 4!

6! = 6 Ã— 5!

So by using these values

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

1/4! + 1/(5Ã—4!) = x/(6Ã—5!)

(5 + 1) / (5Ã—4!) = x/(6Ã—5!)

6/5! = x/(6Ã—5!)

x = (6 Ã— 6 Ã— 5!)/5!

= 36

âˆ´ x = 36.

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

We know that

10! = 10 Ã— 9!

9! = 9 Ã— 8!

Using these values, we get

x/10! = 1/8! + 1/9!

x/10! = 1/8! + 1/(9Ã—8!)

x/10! = (9 + 1) / (9Ã—8!)

x/10! = 10/9!

x/(10Ã—9!) = 10/9!

x = (10 Ã— 10 Ã— 9!)/9!

= 10 Ã— 10

= 100

âˆ´ x = 100.

(iii) 1/6! + 1/7! = x/8!

We know that

8! = 8 Ã— 7 Ã— 6!

7! = 7 Ã— 6!

So by using these values,

1/6! + 1/7! = x/8!

1/6! + 1/(7Ã—6!) = x/8!

(1 + 7)/(7Ã—6!) = x/8!

8/7! = x/8!

8/7! = x/(8Ã—7!)

x = (8 Ã— 8 Ã— 7!)/7!

= 8 Ã— 8

= 64

âˆ´ The value of x is 64.

### (iv) 1 â‹… 3 â‹… 5 â‹… 7 â‹… 9 â€¦ (2n â€“ 1)

Solution:

(i) 5 â‹… 6 â‹… 7 â‹… 8 â‹… 9 â‹… 10

We can rewrite the above expression as:

5 â‹… 6 â‹… 7 â‹… 8 â‹… 9 â‹… 10 = (1Ã—2Ã—3Ã—4Ã—5Ã—6Ã—7Ã—8Ã—9Ã—10)/(1Ã—2Ã—3Ã—4)

= 10!/4!

(ii) 3 â‹… 6 â‹… 9 â‹… 12 â‹… 15 â‹… 18

We can rewrite the above expression as:

3 â‹… 6 â‹… 9 â‹… 12 â‹… 15 â‹… 18 = (3Ã—1) Ã— (3Ã—2) Ã— (3Ã—3) Ã— (3Ã—4) Ã— (3Ã—5) Ã— (3Ã—6)

= 36 (1Ã—2Ã—3Ã—4Ã—5Ã—6)

= 36 (6!)

(iii) (n + 1) (n + 2) (n + 3) â€¦ (2n)

We can rewrite the above expression as:

(n + 1) (n + 2) (n + 3) â€¦ (2n) = [(1) (2) (3) ..(n) â€¦ (n + 1) (n + 2) (n + 3) â€¦ (2n)] / (1) (2) (3) .. (n)

= (2n)!/n!

(iv) 1 â‹… 3 â‹… 5 â‹… 7 â‹… 9 â€¦ (2n â€“ 1)

We can rewrite the above expression as:

1 â‹… 3 â‹… 5 â‹… 7 â‹… 9 â€¦ (2n â€“ 1) = [(1) (3) (5) â€¦ (2n-1)] [(2) (4) (6) â€¦ (2n)] / [(2) (4) (6) â€¦ (2n)]

Take 2 out from each term of denominator

= [(1) (2) (3) (4) â€¦ (2n-1) (2n)] / 2n [(1) (2) (3) â€¦ (n)]

= (2n)! / 2nn!

### (ii) (2 Ã— 3)! = 2! Ã— 3!

Solution:

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

LHS:

(2 + 3)! = 5!

RHS,

2! + 3! = (2Ã—1) + (3Ã—2Ã—1)

= 2 + 6

= 8

LHS â‰  RHS

âˆ´ The given expression is false.

(ii) (2 Ã— 3)! = 2! Ã— 3!

LHS:

(2 Ã— 3)! = 6!

= 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1

= 720

RHS,

2! Ã— 3! = (2Ã—1) Ã— (3Ã—2Ã—1)

= 12

LHS â‰  RHS

âˆ´ 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 Ã— 4 Ã— 3

On comparing both sides, we get

n = 3

âˆ´ 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 Ã— 9

On comparing both sides,we get

n = 9

âˆ´ 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 Ã— 7

On comparing both sides,we get

n + 2 = 7

âˆ´ 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)! Ã— 2!(n – 2)!] / [3!(2n – 3)! Ã— n!] = 44/3

[2nÃ—(2n-1)Ã—(2n-2)Ã—(2n-3)!Ã—2!Ã—(n-2)!] / [3!Ã—(2n-3)!Ã—nÃ—(n-1)Ã—(n-2)!] = 44/3

(2nÃ—(2n-1)Ã—2(n-1))/(3Ã—nÃ—(n-1)) = 44/3

4(2n-1) = 44

2n-1 = 11

2n = 12

âˆ´ n = 6

### 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)!  = [nÃ—(n-1)Ã—(n-2) . . . (n-r+2)Ã—(n-r+1)Ã—(n-r)!] / (n-r)!

= nÃ—(n-1)Ã—(n-2). . . (n-r+2)Ã—(n-r+1)

= nÃ—(n-1)Ã—(n-2). . . (n-(r-2))Ã—(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! = [1Ã—2Ã—3Ã—4 . . . (2n-2)Ã—(2n-1)Ã—(2n)Ã—(2n+1)] / n!

Separate even and odd terms in numerator

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

Take factor of two out from all even terms of numerator

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

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

= 2n [1Ã—3Ã—5. . . (2n-1)Ã—(2n+1)]

= RHS

LHS = RHS

Hence, proved.

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