Question

Question 2 Determine n if

(i) 2nC3 : nC3 = 12 : 1

(ii) 2nC3 : nC3 = 11 : 1

Class X1 - Maths -Permutations and Combinations Page 153

Answer 1

(i) ²ⁿC₃ : ⁿC₂ = 12 : 1

use the concept, ⁿCₐ =n!/a!(n-a)!

2n(2n-1)(2n-2)(2n-3)!/3!(2n-3)! : n(n-1)(n-2)!/2!(n-2)! = 12 : 1

{2n(2n -1)(2n-1)/6}/{n(n-1)/2} = 12/1

{4n(2n-1)(n-1)}/3{n(n-1)}=12

4(2n-1)/3 = 12

2n - 1 = 9

2n = 10

n = 5

(ii) ²ⁿC₃ : ⁿC₃ = 11 : 1

{2n(2n-1)(2n-2)(2n-3)!/3!(2n-3)!}/{n(n-1)(n-2)(n-3)!/3!(n-3)!} = 11 / 1

{ 4n(n-1)(2n-1)/6}/{n(n-1)(n-2)/6} = 11

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

8n - 4 = 11n -22

8n -11n = -22+ 4

-3n = -18

n = 6

Answer 2

Answer:

Step-by-step explanation:

2nC3:nC3=12:1

2nC3÷nC3=12

(2n(2n-1)(2n-2)÷6)÷(12×n(n-1)(n-2)÷6)

2(n-1)(n-1)=6(n-1)(n-2)

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

2n-1=3n-6

-1+6=3n-2n

n=5