Permutations and Combinations Class 11 OP Malhotra Exe-12B ISC Maths Solutions Ch-12 Solutions. In this article you would learn about Permutations (nPr) and Factorial Notation. 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-12B 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(B)	Permutations (nPr) and Factorial Notation.

Exercise- 12B

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

Que-1: Evaluate 4! / (2!2!).

Sol: 4! / (2!2!)
= (4×3×2×1)/(2×1×2×1)
= 24/4 = 6

Que-2: Give the meaning and value of the symbol in the following :
(i) ⁵P₂
(ii) ⁷P₃
(iii) ¹⁰P₄

Sol: (i) 5P2 = 5! / (5−2)!
= 5!/3! = (5×4×3!)/3!
= 20

(ii) 7P3 = 7! / (7−3)! = 7!/4!
= (7×6×5×4!)/4! = 210

(iii) 10P4 = 10! / (10−4)! = 10!/6!
= (10×9×8×7×6!)/6! = 5040

Find n if :

Que-3: nP2 = 30

Sol: Given ⁿP₂ = 30
⇒ n! / (n−2)! = 30
⇒ {n(n−1)(n−2)!} / (n−2)! = 30
⇒ n² – n – 30 = 0
⇒ (n – 6) (n + 5) = 0
⇒ n = 6, – 5
But n ∈ N
∴ n = 6

Que-4: ⁿP₄ : ^{n – 1}P₃ = 9 : 1

Sol: Given : ⁿP₄ : ^{n – 1}P₃ = 9 : 1

n = 9

Que-5: ²ⁿP_{n + 1} : ^{2n – 2}P_n = 56 : 3

Sol: Given : ²ⁿP_{n + 1} : ^{2n – 2}P_n = 56 : 3

⇒ 3n (2n – 1) = 28(n – 1)
⇒ 6n² – 31n + 28 = 0
⇒ (n – 4)(6n – 7) = 0
⇒ n = 4, 7/6 but n ∈ N
∴ n = 4

Que-6: ^{2n + 1}P_{n – 1} : ^{2n – 1}P_n = 3 : 5

Sol: Given : ^{2n + 1}P_{n – 1} : ^{2n – 1}P_n = 3 : 5

⇒ 10(2n + 1) = 3(n² + 3n + 2)
⇒ 3n² – 11n – 4 = 0
⇒ (n – 4) (3n + 1) = 0
⇒ n = 4, –1/3 but n ∈ N
∴ n = 4

Que-7: ²ⁿP₃ = 100. ⁿP₂ …(1)

Sol: ⇒ 2n! / (2n−3)! = 100 × n! / (n−2)!
⇒ 2n(2n – 1) (2n – 2) = 100n (n – 1)
⇒ 4n (n – 1) [2n – 1 – 25] = 0
⇒ n = 0, 1, 13
but n = 0, 1 does not satisfies eqn. (1)
∴ n = 13

Que-8: P (n, 6) = 3P (n, 5)

Sol: Given P (n, 6) = 3P (n, 5)
⇒ ⁿP₆ = 3 × ⁿP₅
⇒ n! / (n−6)! = 3 × [n! / (n−5)!]
⇒ {(n−5)(n−6)!} / (n−6)! = 3
⇒ n – 5 = 3 ⇒ n = 8

Que-9: 2P (n, 3) = P (n + 1, 3)

Sol: Given 2P (n, 3) = P (n + 1, 3)
⇒ 2 ⁿP₃ = ^{n + 1}P₃
⇒ 2 . [n! / (n−3)!] = (n+1)! / (n−2)!
⇒ 2 . [n! / (n−3)!] = {(n+1)n!} / {(n−2)(n−3)!}
⇒ n + 1 = 2 (n – 2) ⇒ n = 5

Que-10: Find r if 5P (4, r) = 6P (5, r – 1), r ≥ 1.

Sol: Given 5P (4, r) = 6P (5, r- 1), r ≥ 1
⇒ 4 . ⁴P_r = 6 . ⁵P_{r – 1}

⇒ (6 – r) (5 – r) = 6 ⇒ r² – 11r + 24 = 0
⇒ (r – 3)(r – 8) = 0 ⇒ r = 3, 8
When r = 8 Then ⁴P_r = ⁴P₈ is meaningless
Therefore r = 3

Prove that:

Que-11: P (n, n) = 2P (n, n – 2)

Sol: L.H.S. = p (n, n) = nPn = n! / (n−n)! = n! / 0! = n!
R.H.S = 2P (n, n – 2) = 2 × n P n – 2
= 2 × [n! / (n−n+2)!] = 2 x (n!/2!) = n!
∴ L.H.S = R.H.S

Que-12: P (10, 3) = P (9, 3) + 3P (9, 2)

Sol: L.H.S = P (10, 3) = 10P3
= 10! / (10−3)! = 10! / 7!
= (10×9×8×7!)/7! = 720
R.H.S = P (9, 3) + 3P (9, 2) = 9P3 + 3 . 9P2
= 9! / (9−3)! + 3 .9! / (9−2)! = (9!/6!) + 3. (9!/7!)
= {(9×8×7×6!)/6!} + {(3×9×8×7!)/7!}
= 9 × 8 × 7 + 3 × 9 × 8
= 9 × 8(7 + 3) = 9 × 8 × 10 = 720
∴ L.H.S = R.H.S

Que-13: P (n, r) = (n – r + 1) P (n, r – 1)

Sol: L.H.S. = P(n, r) = nPr = n! / (n−r)!
R.H.S. = (n – r + 1) P (n, r + 1)
= (n – r + 1) {n! / (n−r+1)!}
= {(n−r+1)n!} / [(n−r+1)(n−r)!]
= n! / (n−r)!
Thus L.H.S. = R.H.S.

Que-14: P(n, n) = P (n, n – 1)

Sol: L.H.S = P(n, n) = nPn
= n! / (n−n)! = n!/0! = n!
and R.H.S = p (n, n – 1) = n P (n – 1)
= n! / (n−n+1)! = n!/1! = n!
Thus, L.H.S = R.H.S

Que-15: If (1/9!) + (1/10!) = (x/11!), find x.

Sol: Given (1/9!) + (1/10!) = (x/11!)
⇒ x = (11/9!) + (11/10!)
= {(11×10×9!)/9!} + {(11×10!)/10!}
⇒ x = 110 + 11 = 121

Que-16: If {n!/2!(n−2)!} and {n!/4!(n−4)!} are in the ratio 2 : 1, find the value of n.

Sol: {n!/2!(n−2)!} : {n!/4!(n−4)!} = 2 : 1 …(1)
⇒ {n!/2!(n−2!)} / {n!/4!(n−4)!} = 2/1
⇒ {4!(n−4)!} / {2!(n−2)!} = 2
⇒ 12 / {(n−2)(n−3)} = 2
⇒ 6 = n² – 5n + 6
⇒ n = 0, 5
but n = 0 does not satisfies eqn. (1)
∴ n = 5

Solve for n:

Que-17: (2n)! / {3!(2n−3)!} : n!/ {2!(n−2)!} = 44 : 3

Sol: Given
[(2n)!/{3!(2n−3)!}] : [n!/{2!(n−2)!}] = 44 : 3 …(1)
⇒ [2n!/{3!(2n−3)!}] / [n!/{2!(n−2)!}] = 44/3
⇒ [{2n(2n−1)(2n−2)}/3!] × [2!/{n(n−1)}] = 44/3
⇒ n(2n – 1) (n – 1) = 11n (n – 1)
⇒ n(n -1) [2n – 1 – 1 – 1] = 0
⇒ n = 0, 1, 6
but n = 0, 1 does not satisfies eqn. (1)
∴ n = 6

Que-18: (n + 1) ! = 56.(n – 1) !

Sol: Given (n + 1) ! = 56(n – 1)!
⇒ (n + 1) n(n – 1)! = 56(n – 1)!
⇒ n² + n – 56 = 0
⇒ (n – 7)(n + 8) = 0
⇒ n = 7,-8 but n ∈ N
∴ n = 7

Que-19: Prove that 2n!/n! = 1 . 3. 5 … (2n – 1) 2ⁿ.

Sol: LHS = 2n!/n!

Que-20: Convert into factorial : 7 . 8 . 9 . 10 . 11 . 12 . 13 . 14 . 15

Sol: 7 . 8 . 9 . 10 . 11 . 12 . 13 . 14 . 15
= (1⋅2⋅3⋅4⋅5⋅6⋅7⋅8⋅9⋅10⋅11⋅12⋅13⋅14⋅15)/(1⋅2⋅3⋅4⋅5⋅6)
= 15! / 6!

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

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