Question

2nC4:nC3=35:2 find n​

Answer 1

Answer:

n = 8

Step-by-step explanation:

I've explained it in the above attached picture.

Hope it helps you. :)

Answer 2

n= 4

Explanation:

Formula :  $^nC_r=\dfrac{n!}{r!(n-r)!}$

Then,

$^{2n}C_4=\dfrac{2n!}{4!(2n-4)!}\\\\=\dfrac{2n(2n-1)(2n-2)(2n-3)(2n-4)!}{4!(2n-4)!}\\\\\\=\dfrac{2n(2n-1)(2n-2)(2n-3)}{4!}$

$^nC_3=\dfrac{n!}{3!(n-3)!}=\dfrac{n(n-1)(n-2)(n-3)!}{3!(n-3)!}\\\\=\dfrac{n(n-1)(n-2)}{3!}$

Given : $^{2n}C_4:\ ^nC_3=35:2$

i.e.  

$\dfrac{^{2n}C_4}{^nC_3}=\dfrac{35}{2}\\\\\\\dfrac{\dfrac{2n(2n-1)(2n-2)(2n-3)}{4!}}{\dfrac{n(n-1)(n-2)}{3!}}=\dfrac{35}{2}\\\\\\\dfrac{2n(2n-1)(2n-2)(2n-3)}{4\times3!}\times\dfrac{3!}{n(n-1)(n-2)}=\dfrac{35}{2}\\\\\\\dfrac{2n(2n-1)(2)(n-1)(2n-3)}{4(n(n-1)(n-2))}=\dfrac{35}{2}\\\\\\\dfrac{(4)(2n-1)(2n-3)}{4(n-2)}=\dfrac{35}{2}\\\\\\\dfrac{(2n-1)(2n-3)}{(n-2)}=\dfrac{35}{2}$

$(2)(2n-1)(2n-3)=35(n-2)\\\\ 8 n^2 - 51 n + 76 = 0\\\\(n-4)(n-\dfrac{19}{8})=0\\\\ n= 4\ or\ n=\dfrac{19}{8}$

 

But n is a natural number .

∴ n cannot be $\dfrac{19}{8}$ .

Therefore , n= 4

Learn more :

If nCr - 1 = 36,nCr = 84 and nCr+1 = 126, then find rC2. [NT : form eqn using nCr/nCr+1 and nCr/nCr-1 to find the value of r.]

https://brainly.in/question/3343721  [Answered by  Abhi178]