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

Effective OP Malhotra Maths Class 12 Solutions Chapter 12 Permutations and Combinations Ex 12(d) can help bridge the gap between theory and application.

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

Question 1.
Find the value of:
(i) ⁵C₂
(ii) ¹⁰C₄
(iii) ⁵⁰C₄₇
Solution:
(i) ⁵C₂ = \(\frac{5 !}{(5-2) ! 2 !}\) = \(\frac{5 !}{3 ! 2 !}\)
= \(\frac{5 \times 4 \times 3 !}{3 ! \times 2}\) = 10

(ii) ¹⁰C₄ = \(\frac{10 !}{(10-4) ! 4 !}\) = \(\frac{10 !}{6 ! 4 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 !}{6 ! \times 4 \times 3 \times 2 \times 1}\) = 210

(iii) ⁵⁰C₄₇ = \(\frac{50 !}{(50-47) ! 47 !}\) = \(\frac{50 !}{3 ! 47 !}\)
= \(\frac{50 \times 49 \times 48 \times 47 !}{6 \times 47 !}\) = 19600

Question 2.
Evaluate :
(i) C(15, 14)
(ii) C(8, 5)
(iii) ¹¹C₂
Solution:
(i) C(15, 14) = ¹⁵C₁₄ = \(\frac{15 !}{1 ! 14 !}\) = 15
(ii) C(8, 5) = ⁸C₅ = \(\frac{8 !}{3 ! 5 !}\) = \(\frac{8 \times 7 \times 6 \times 5 !}{6 \times 5 !}\) = 56
(iii) ¹¹C₂ = \(\frac{11 !}{2 ! 9 !}\) = \(\frac{11 \times 10 \times 9 !}{2 ! \times 9 !}\) = 55

Question 3.
Evaluate:
3. Evaluate :
(i) C(19, 17) + C(19, 18)
(ii) C(31, 26) – C(30, 26)
Solution:
(i) C (19, 17) + C(19, 18) = ¹⁹C₁₇ + ¹⁹C₁₈
= \(\frac{19 !}{2 ! 17 !}\) + \(\frac{19 !}{1 ! 18 !}\)
= \(\frac{19 \times 18}{2}\) + 19
= 171 + 19 = 190

(ii) C(31, 26) – C (30, 26) = ³¹C₂₆ – ³⁰C₂₆
= \(\frac{31 !}{5 ! 26 !}\) – \(\frac{30 !}{4 ! 26 !}\)
= \(\frac{31 \cdot 30 !}{5 \cdot 4 ! 26 !}\) – \(\frac{30 !}{4 ! 26 !}\)
= \(\frac{30 !}{26 ! 4 !}\) \(\left[\frac{31}{5}-1\right]\)
= \(\frac{30 \times 29 \times 28 \times 27}{4 \times 3 \times 2 \times 1}\) × \(\frac{26}{5}\) = 142506

Question 4.
If ⁴C₂ = n ⁴C₂, find n.
Solution:
Given ⁴C₂ = n. ⁴C₂ ⇒ \(\frac{4 !}{2 !}\) = n . \(\frac{4 !}{2 ! 2 !}\)
⇒ n = 2! = 2

Question 5.
If ⁿC₄ = ⁿC₆, find n.
Solution:
Given ⁿC₄ = ⁿC₆ ⇒ n = 4 + 6 = 10
[if ⁿC_r = ⁿC_s Then r = s or r + s = n]

Question 6.
If C (2n, 3) : C(n, 2) = 12 : 1, find n.
Solution:
Given C (2n, 3) : C(n, 2) = 12 : 1 …(1)
⇒ \(\frac{{ }^{2 n} C_3}{{ }^n C_2}\) = \(\frac{12}{1}\)
⇒ \(\frac{2 n(2 n-1)(2 n-2)}{3 n(n-1)}\) = \(\frac{12}{1}\)
⇒ 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

Question 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.
Solution:
Given ⁿC_r : ⁿC_r+1 = 1 : 2
⇒ \(\frac{\frac{n !}{(n-r) ! r !}}{\frac{n !}{(n-r-1) !(r+1) !}}\) = \(\frac{1}{2}\)
⇒ \(\frac{n !}{(n-r) ! r !}\) × \(\frac{(n-r-1) !(r+1) !}{n !}\) = \(\frac{1}{2}\)
⇒ \(\frac{r+1}{n-r}\) = \(\frac{1}{2}\)
⇒ 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

Question 8.
If C(n, 10) = C(n, 12), determine C(n, 5).
Solution:
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
= \(\frac{22 !}{17 ! 5 !}\)
= \(\frac{22 \times 21 \times 20 \times 19 \times 18}{5 \times 4 \times 3 \times 2}\)
= 26334

Question 9.
If C (2n, r) = C(2n, r + 2), find r in terms of n.
Solution:
Given C(2n, r) = C(2n, r + 2)
⇒ ²ⁿC_r = ²ⁿC_r+2
⇒ \(\frac{2 n !}{(2 n-r) ! r !}\) = \(\frac{2 n !}{(2 n-r-2) !(r+2) !}\)
⇒ \(\frac{(r+2) !}{r !}\) = \(\frac{(2 n-r) !}{(2 n-r-2) !}\)
⇒ (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 = \(\frac{(2 r+1) \pm \sqrt{(2 r+1)^2+8(r+1)}}{4}\)
⇒ n = \(\frac{(2 r+1) \pm \sqrt{(2 r+3)^2}}{4}\)
⇒ n = \(\frac{2 r+1 \pm 2 r+3}{4}\) = \(\frac{4 r+4}{4}\), \(\)
since n can’t be fractional ∴ n = r + 1

Question 10.
The value of \({ }^{50} \mathrm{C}_4+\sum_{r=1}^6{ }^{56-r} \mathrm{C}_3\) is
(a) ⁵⁵C₄
(b) ⁵⁵C₃
(c) ⁵⁶C₃
(d) ⁵⁶C₄
Solution:
\({ }^{50} \mathrm{C}_4+\sum_{r=1}^6{ }^{56-r} \mathrm{C}_3\) = 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)

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