Permutations and Combinations Class 11 OP Malhotra Exe-12D ISC Maths Solutions Ch-12 Solutions. In this article you would learn about Combination (nCr) and its Properties. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.

Permutations and Combinations Class 11 OP Malhotra Exe-12D ISC Maths Solutions Ch-12

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-12	Permutations and Combinations
Writer	O.P. Malhotra
Exe-12(D)	Combination (nCr) and its Properties.

Exercise- 12D

Permutations and Combinations Class 11 OP Malhotra Exe-12D Solution.

Que-1: Find the value of:
(i) ⁵C₂
(ii) ¹⁰C₄
(iii) ⁵⁰C₄₇

Sol:  (i) 5C2 = 5! / {(5−2)!2!} = 5!/(3!.2!)
= (5×4×3!)(3!×2) = 10

(ii) 10C4 = 10! / {(10−4)!4!} = 10!/(6!.4!)
= (10×9×8×7×6!)/(6!×4×3×2×1) = 210

(iii) 50C47 = 50! / {(50−47)!47!} = 50!/(3!.47!)
= (50×49×48×47!)/(6×47!) = 19600

Que-2: Evaluate :
(i) C(15, 14)
(ii) C(8, 5)
(iii) ¹¹C₂

Sol: (i) C(15, 14) = 15C14 = 15!/(1!.14!) = 15

(ii) C(8, 5) = 8C5 = (8!/3!.5!) = (8×7×6×5!)/(6×5!) = 56

(iii) 11C2 = (11!/(2!.9!) = (11×10×9!)/(2!×9!) = 55

Que-3: Evaluate:
(i) C(19, 17) + C(19, 18)
(ii) C(31, 26) – C(30, 26)

Sol: (i) C (19, 17) + C(19, 18) = ¹⁹C₁₇ + ¹⁹C₁₈
= {19!/(2!.17!)} + {19!/(1!.18!)}
= {(19×18)/2} + 19
= 171 + 19 = 190

(ii) C(31, 26) – C (30, 26) = ³¹C₂₆ – ³⁰C₂₆
= {31!/(5!.26!)} – {30!/(4!.26!)}
= {(31⋅30!)/(5⋅4!26!)} – {30!/(4!.26!)}
= {30!/(26!.4!)} [(31/5)−1]
= (30×29×28×27)/(4×3×2×1) × (26/5) = 142506

Que-4: If ⁴P₂ = n ⁴C₂, find n.

Sol: Given ⁴C₂ = n. ⁴C₂
⇒ 4!/2! = n . {4!/(2!.2!)}
⇒ n = 2! = 2

Que-5: If ⁿC₄ = ⁿC₆, find n.

Sol: Given ⁿC₄ = ⁿC₆ ⇒ n = 4 + 6 = 10
[if ⁿC_r = ⁿC_s Then r = s or r + s = n]

Que-6: If C (2n, 3) : C(n, 2) = 12 : 1, find n.

Sol: Given C (2n, 3) : C(n, 2) = 12 : 1 …(1)
⇒ (2nC3) / (nC2) = 12/1
⇒ {2n(2n−1)(2n−2)} / {3n(n−1)} = 12/1
⇒ n(2n – 1)(n – 1) = 9n(n – 1)
⇒ n(n – 1)[2n – 1 – 9] = 0
⇒ n = 0, 1, 5
but n = 0, 1 does not satisfies eqn. (1).
Thus n = 5

Que-7: If ⁿC_r : ⁿC_r+1 = 1 : 2
and ⁿC_r+1 : ⁿC_r+2 = 2 : 3
determine the values of n and r.

Sol: Given nCr : nCr+1 = 1 : 2
⇒ [n!/{(n−r)!r!}] / [n!/{(n−r−1)!(r+1)!}] = 1/2
⇒ [n!/{(n−r)!r!}] × [{(n−r−1)!(r+1)!}/n!] = 1/2
⇒ (r+1)/(n−r) = 1/2
⇒ 2r + 2 = n – r
⇒ n – 3r = 2 …(1)
and nC(r+1) : nC(r+2) = 2 : 3

⇒ 3r + 6 = 2n – 2r -2
⇒ 2n – 5r = 8 …(2)
eqn. (2) – 2 eqn. (1); we have
r = 8 – 4 = 4
∴ from (1) ; n – 12 = 2 ⇒ n = 14

Que-8: If C(n, 10) = C(n, 12), determine C(n, 5).

Sol: Given C(n, 10) = C(n, 12)
⇒ ⁿC₁₀ = ⁿC₁₂
⇒ n = 10 + 12 = 22
[if ⁿC_r = ⁿC_s ⇒ r = s ⇒ r + s = n]
∴ C(n, 5) = ⁿC₅ = ²²C₅
= 22!/(17!.5!)
= (22×21×20×19×18)/(5×4×3×2)
= 26334

Que-9: If C (2n, r) = C(2n, r + 2), find r in terms of n.

Sol: Given C(2n, r) = C(2n, r + 2)
⇒ ²ⁿC_r = ²ⁿC_r+2
⇒ {2n!/{(2n−r)!r!}} = [2n!/{(2n−r−2)!(r+2)!}]
⇒ [(r+2)!/r!] = [(2n−r)!/(2n−r−2)!]
⇒ (r + 2)(r + 1) = (2n – r)(2n – r – 1)
⇒ r² + 3r + 2 = 4n² + r² -4nr – 2n + r
⇒ 4n² – 4nr – 2n – 2r – 2 = 0
⇒ 2n² – 2nr – n – r – 1 =0
⇒ 2n² – n(2r + 1) – r – 1 = 0
⇒ n = [(2r+1)±{√((2r+1)²+8(r+1))}]/4
⇒ n = [(2r+1)±√(2r+3)²]/4
⇒ n = (2r+1±2r+3)/4 = (4r+4)/4,
since n can’t be fractional ∴ n = r + 1

Que-10: The value of 50C4 + {∑6(r=1) (56−r)C3 is
(a) ⁵⁵C₄
(b) ⁵⁵C₃
(c) ⁵⁶C₃
(d) ⁵⁶C₄

Sol: 50C4 + {∑6(r=1) (56−r)C3
= ⁵⁰C₄ + ⁵⁵C₃ + ⁵⁴C₃ + ⁵³C₃ + ⁵²C₃ +⁵¹C₃ + ⁵⁰C₃
= (⁵⁰C₃ + ⁵⁰C₄) + ⁵¹C₃ + ⁵²C₃ + ⁵³C₃ + ⁵⁴C₃ + ⁵⁵C₃ [∵ ⁿC_r + ⁿC_{r – 1} = ⁿ⁺¹C_r]
= (⁵¹C₄ + ⁵¹C₃) + ⁵²C₃ + ⁵³C₃ + ⁵⁴C₃ + ⁵⁵C₃
= (⁵²C₄ + ⁵²C₃) + ⁵³C₃ + ⁵⁴C₃ + ⁵⁵C₃
= (⁵³C₄ + ⁵³C₃) + ⁵⁴C₃ + ⁵⁵C₃
= (⁵⁴C₄ + ⁵⁴C₃) + ⁵⁵C₃
= ⁵⁵C₄ + ⁵⁵C₃ = ⁵⁶C₄
∴ Ans (d)

Que-11: ^{n – 1}C₃ + ^{n – 1}C₄ > ⁿC₃ if
(a) n > 7
(b) n ≥ 7
(c) n > 6
(d) n ≥ 6

Sol: Given : ^{n – 1}C₃ + ^{n – 1}C₄ > ⁿC₃

Ans (a)

–: End of Permutation and Combinations Class 11 OP Malhotra Exe-12D ISC Maths Ch-12 Solutions. :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions

Thanks

Please share with your friends