The least value of n so that C 3 n - 1 + C 4 n - 1 > C 3 n

Given:

${}^{\left(n-1\right)}C_3+{}^{\left(n-1\right)}C_4>{}^{n}C_3$

$\Rightarrow {}^{\left(n-1\right)}C_{4-1}+{}^{\left(n-1\right)}C_4>{}^{n}C_3$

$\Rightarrow {}^{n}C_4>{}^{n}C_3$

$\left[\because {}^{n}C_{r-1}+{}^{n}C_r={}^{n+1}C_r\right]$

$\Rightarrow \frac{n!}{4!\times \left(n-4\right)!}>\frac{n!}{3!\times \left(n-3\right)!}$

$\Rightarrow 3!\times \left(n-3\right)!>4!\times \left(n-4\right)!$

$\Rightarrow 3!\times \left(n-3\right)\times \left(n-4\right)!>4\times 3!\times \left(n-4\right)!$

$\Rightarrow \left(n-3\right)>4$

$\Rightarrow n>7$

So, least value of $n$ is $8$.