Determine n if,
(i) ²ⁿC₃ : ⁿC₃ = 12 : 1   (ii) ²ⁿC₃ : ⁿC₃= 11 : 1

Solution:

(i) It is given that  ²ⁿC₃/ⁿC₃ = 12/1

Using nCr formula,

(2n)! / [3!(2n - 3)!] ×  [3!(n - 3)!] / n! = 12

[(2n)(2n-1)(2n-2)(2n-3)!] / [6 (2n-3)!] × [6 (n-3)! / [n(n-1)(n-2)(n-3)!] = 12

[(2)(2n - 1)(2n - 2)] / [(n - 1)(n - 2)] = 12

[4(2n - 1)(n - 1)] / [(n - 1)(n - 2)] = 12

(2n - 1)/(n - 2) = 3

2n - 1 = 3n - 6

-n + 5 = 0

n = 5
 

(ii) ²ⁿC₃/ⁿC₃ = 11/1

Using nCr formula,

(2n)! / [3!(2n - 3)!] ×  [3!(n - 3)!] / n! = 11

[(2n)(2n-1)(2n-2)(2n-3)!] / [6 (2n-3)!] × [6 (n-3)! / [n(n-1)(n-2)(n-3)!] = 11

[2(2n - 1)(2n - 2)] / [(n - 1)(n - 2)] = 11

[4(2n - 1)(n - 1)] / [(n - 1)(n - 2)] = 11

[(2n - 1)(n - 1)] / [(n - 1)(n - 2)] = 11/4

(2n - 1)/(n - 2) = 11/4

4(2n - 1) = 11(n - 2)

8n - 4 = 11n - 22

8n - 11n = 4 - 22

-3n = -18

n = 6

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NCERT Solutions Class 11 Maths Chapter 7 Exercise 7.4 Question 2

Determine n if, (i) ²ⁿC₃ : ⁿC₃ = 12 : 1 (ii) ²ⁿC₃ : ⁿC₃= 11 : 1

Summary:

(i) If ²ⁿC₃ : ⁿC₃ = 12 : 1, then n = 5 (ii) If ²ⁿC₃ : ⁿC₃ = 11 : 1 then n = 6