Question

Q67. Determine n if ²ⁿC₃ : ⁿC₃ = 11 : 1

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Answer 1

ANSWER:

Given:

• ²ⁿC₃ : ⁿC₃ = 11 : 1

To Find:

• Value of n

Solution:

$:\longrightarrow^{2n}\!C_3:^{n}\!C_3=11:1\\\\\text{We know that, ^n\!C_r=\dfrac{n!}{r!(n-r)!} . So,}\\\\:\implies\dfrac{(2n)!}{3!(2n-3)!}:\dfrac{(n)!}{3!(n-3)!} = 11:1\\\\:\implies\dfrac{\dfrac{(2n)!}{3!\!\!\!/\:(2n-3)!}}{\dfrac{(n)!}{3!\!\!\!/\:(n-3)!}}=\dfrac{11}{1}\\\\:\implies\dfrac{\dfrac{(2n)!}{(2n-3)!}}{\dfrac{(n)!}{(n-3)!}}=11\\\\\text{On cross-multiplying,}\\\\:\implies\dfrac{(2n)!}{(2n-3)!}=11\times\dfrac{(n)!}{(n-3)!}\\\\:\implies\dfrac{2n(2n-1)(2n-2)(2n-3)!}{(2n-3)!}=11\times\dfrac{n(n-1)(n-2)(n-3)!}{(n-3)!}\\\\:\implies2n\!\!\!/(2n-1)(2n-2)=11(n\!\!\!/)(n-1)(n-2)\\\\:\implies2(2n-1)2(n-1)=11(n-1)(n-2)\\\\:\implies4(2n-1)=11(n-2)\\\\:\implies8n-4=11n-22\\\\:\implies11n-8n=22-4\\\\:\implies3n=18\\\\\bf{:\implies n=6}$

Formula Used:

• $^n\!C_r=\dfrac{n!}{r!(n-r)!}$