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RD Chapter 16- Permutations Ex-16.1 Interview Questions Answers

Question 1 :
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!

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

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.

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!

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.

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)

Answer 4 :

(i)  6  7  8  9  10

Let us evaluate

We can write it as:

 6  7  8  9  10 = (1×2×3×4×5×6×7×8×9×10)/(1×2×3×4)

= 10!/4!

(ii)  6  9  12  15  18

Let us evaluate

 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)  3  5  7  9 … (2n – 1)

Let us evaluate

 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!

Question 5 :
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.

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

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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RD Chapter 16- Permutations Ex-16.1 Contributors

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