OP Malhotra Class 11 Maths Solutions Chapter 12 Permutations and Combinations Ex 12(b)

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S Chand Class 11 ICSE Maths Solutions Chapter 12 Permutations and Combinations Ex 12(b)

Question 1.
Evaluate \(\frac{4 !}{2 ! 2 !}\).
Solution:
\(\frac{4 !}{2 ! 2 !}\) = \(\frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1}\) = \(\frac{24}{4}\) = 6

Question 2.
Give the meaning and value of the symbol in the following :
(i) ⁵P₂
(ii) ⁷P₃
(iii) ¹⁰P₄
Solution:
(i) ⁵P₂ = \(\frac{5 !}{(5-2) !}\) = \(\frac{5 !}{3 !}\) = \(\frac{5 \times 4 \times 3 !}{3 !}\) = 20

(ii) ⁷P₃ = \(\frac{7 !}{(7-3) !}\) = \(\frac{7 !}{4 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 !}{4 !}\) = 210

(iii) ¹⁰P₄ = \(\frac{10 !}{(10-4) !}\) = \(\frac{10 !}{6 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 !}{6 !}\) = 5040

Question 3.
Find n if:
Solution:
Given ⁿP₂ = 30
⇒ \(\frac{n !}{(n-2) !}\) = 30
⇒ \(\frac{n(n-1)(n-2) !}{(n-2) !}\) = 30
⇒ n² – n – 30 = 0
⇒ (n – 6) (n + 5) = 0
⇒ n = 6, – 5
But n ∈ N
∴ n = 6

Question 4.
ⁿP₄ : ^{n – 1}P₃ = 9 : 1
Solution:
Given ⁿP₄ : ^{n – 1}P₃ = 9 : 1

Question 5.
²ⁿP_{n + 1} : ^{2n – 2}P_n = 56 : 3
Solution:
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, \(\frac { 7 }{ 6 }\) but n ∈ N
∴ n = 4

Question 6.
^{2n + 1}P_{n – 1} : ^{2n – 1}P_n = 3 : 5
Solution:
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, –\(\frac { 1 }{ 3 }\) but n ∈ N
∴ n = 4

Question 7.
²ⁿP₃ = 100. ⁿP₂ …(1)
⇒ \(\frac{2 n !}{(2 n-3) !}\) = 100 × \(\frac{n !}{(n-2) !}\)
⇒ 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

Question 8.
P (n, 6) = 3P (n, 5)
Solution:
Given P (n, 6) = 3P (n, 5)
⇒ ⁿP₆ = 3 × ⁿP₅
⇒ \(\frac{n !}{(n-6) !}\) = 3 × \(\frac{n !}{(n-5) !}\)
⇒ \(\frac{(n-5)(n-6) !}{(n-6) !}\) = 3
⇒ n – 5 = 3 ⇒ n = 8

Question 9.
2P (n, 3) = P (n + 1, 3)
Solution:
Given 2P (n, 3) = P (n + 1, 3)
⇒ 2 ⁿP₃ = ^{n + 1}P₃
⇒ 2 . \(\frac{n !}{(n-3) !}\) = \(\frac{(n+1) !}{(n-2) !}\)
⇒ 2 . \(\frac{n !}{(n-3) !}\) = \(\frac{(n+1) n !}{(n-2)(n-3) !}\)
⇒ n + 1 = 2 (n – 2) ⇒ n = 5

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

Question 11.
P (n, n) = 2P (n, n – 2)
Solution:
L.H.S. = p (n, n) = ⁿP_n = \(\frac{n !}{(n-n) !}\) = \(\frac{n !}{0 !}\) = n!
R.H.S = 2P (n, n – 2) = 2 × ⁿP_{n – 2}
= 2 × \(\frac{n !}{(n-n+2) !}\) = 2 x \(\frac{n !}{2 !}\) = n!
∴ L.H.S = R.H.S

Question 12.
P (10, 3) = P (9, 3) + 3P (9, 2)
Solution:
L.H.S = P (10, 3) = ¹⁰P₃
= \(\frac{10 !}{(10-3) !}\) = \(\frac{10 !}{7 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 !}{7 !}\) = 720
R.H.S = P (9, 3) + 3P (9, 2) = ⁹P₃ + 3 . ⁹P₂
= \(\frac{9 !}{(9-3) !}\) + 3 .\(\frac{9 !}{(9-2) !}\) = \(\frac{9 !}{6 !}\) + 3. \(\frac{9 !}{7 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 !}{6 !}\) + \(\frac{3 \times 9 \times 8 \times 7 !}{7 !}\)
= 9 × 8 × 7 + 3 × 9 × 8
= 9 × 8(7 + 3) = 9 × 8 × 10 = 720
∴ L.H.S = R.H.S

Question 13.
P (n, r) = (n – r + 1) P (n, r – 1)
Solution:
L.H.S. = P(n, r) = ⁿP_r = \(\frac{n !}{(n-r) !}\)
R.H.S. = (n – r + 1) P (n, r + 1)
= (n – r + 1) \(\frac{n !}{(n-r+1) !}\)
= \(\frac{(n-r+1) n !}{(n-r+1)(n-r) !}\)
= \(\frac{n !}{(n-r) !}\)
Thus L.H.S. = R.H.S.

Question 14.
P(n, n) = P (n, n – 1)
Solution:
L.H.S = P(n, n) = ⁿP_n
= \(\frac{n !}{(n-n) !}\) = \(\frac{n !}{0 !}\) = n!
and R.H.S = p (n, n – 1) = ⁿP_{n – 1}
= \(\frac{n !}{(n-n+1) !}\) = \(\frac{n !}{1 !}\) = n!
Thus, L.H.S = R.H.S

Question 15.
If \(\frac{1}{9!}\) + \(\frac{1}{10!}\) = \(\frac{x}{11!}\), find x.
Solution:
Given \(\frac{1}{9!}\) + \(\frac{1}{10!}\) = \(\frac{x}{11!}\)
⇒ x = \(\frac{11}{9!}\) + \(\frac{11}{10!}\)
= \(\frac{11 \times 10 \times 9 !}{9 !}\) + \(\frac{11 \times 10 !}{10 !}\)
⇒ x = 110 + 11 = 121

Question 16.
If \(\frac{n !}{2 !(n-2) !}\) and \(\frac{n !}{4 !(n-4) !}\) are in the ratio 2 : 1, find the value of n.
Solution:
Given
\(\frac{n !}{2 !(n-2) !}\) : \(\frac{n !}{4 !(n-4) !}\) = 2 : 1 …(1)
⇒ \(\frac{\frac{n !}{2 !(n-2) !}}{\frac{n !}{4 !(n-4) !}}\) = \(\frac{2}{1}\)
⇒ \(\frac{4 !(n-4) !}{2 !(n-2) !}\) = 2
⇒ \(\frac{12}{(n-2)(n-3)}\) = 2
⇒ 6 = n² – 5n + 6
⇒ n = 0, 5
but n = 0 does not satisfies eqn. (1)
∴ n = 5
Solve for n:

Question 17.
\(\frac{(2 n) !}{3 !(2 n-3) !}\) : \(\frac{n !}{2 !(n-2) !}\) = 44 : 3
Solution:
Given
\(\frac{(2 n) !}{3 !(2 n-3) !}\) : \(\frac{n !}{2 !(n-2) !}\) = 44 : 3 …(1)
⇒ \(\frac{\frac{2 n !}{3 !(2 n-3) !}}{\frac{n !}{2 !(n-2) !}}\) = \(\frac{44}{3}\)
⇒ \(\frac{2 n(2 n-1)(2 n-2)}{3 !}\) × \(\frac{2 !}{n(n-1)}\) = \(\frac{44}{3}\)
⇒ 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

Question 18.
(n + 1) ! = 56.(n – 1) !
Solution:
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

Question 19.
Prove that \(\frac{2 n !}{n !}\) = 1 . 3. 5 … (2n – 1) 2ⁿ.
Solution:

Question 20.
Convert into factorial : 7 . 8 . 9 . 10 . 11 . 12 . 13 . 14 . 15
Solution:
7 . 8 . 9 . 10 . 11 . 12 . 13 . 14 . 15 = \(\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 \cdot 11 \cdot 12 \cdot 13 \cdot 14 \cdot 15}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}\) = \(\frac{15 !}{6 !}\)