If { }^{2n } C ₃:{ }^{n } C ₃=10: 1 , then ratio left(n^2+3 nright):left(n^2-3 n+4right) is

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6.Permutation and Combination

medium

If ${ }^{2n } C _3:{ }^{n } C _3=10: 1$, then the ratio $\left(n^2+3 n\right):\left(n^2-3 n+4\right)$ is

A

$35: 16$

B

$65: 37$

C

$27: 11$

D

$2: 1$

(JEE MAIN-2023)

Solution

$\frac{{ }^{2 n} C_3}{{ }^{{ }^n} C _3}=10 \Rightarrow \frac{2 n (2 n -1)(2 n -2)}{ n ( n -1)( n -2)}=10$

$n =8$

So $\left(n^2+3 n\right):\left(n^2-3 n+4\right)=2$

Standard 11

Mathematics

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