The least value of n so that ${ }^{n-1}{C}_3{+}^{n-1}{C}_4{>}^n{C}_3$

${ }^{n-1}{C}_3{+}^{n-1}{C}_4{>}^n{C}_3$

${ }^n{C}_4{>}^n{C}_3$

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

$\frac{1}{\left(n-4\right)!4.\cancel{3!}}>\frac{1}{\left(n-3\right).\left(n-4\right)!\cancel{3!}}$

$\frac{1}{4}>\frac{1}{n-3}$

$n-3>4$

$n>7$

Least value of n is 8