This exercise discusses the problems based on the factorial (the continued product of first n natural numbers is called the “n factorial”). The solutions here are designed for students who wish to use the RD Sharma textbook as a reference or guide to clarify their doubts pertaining to the problems and learn the correct steps in solving them. RD Sharma Class 11 Maths Solutions contains this exercise in PDF format, which can be easily downloaded from the links given below.
RD Sharma Solutions for Class 11 Maths Exercise 16.1 Chapter 16 – Permutations
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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)
= 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) 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)] / 2n [(1) (2) (3) … (n)]
= (2n)! / 2n n!
5. Which of the following are 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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