Question -14

For LHS

$^nC_r+^nC_{r-1}=^{n+1}C_r$

Substituting $\displaystyle n=n-1 \quad and \quad r=4$

We get $^nC_4> ^nC3$

Least integer $n$ satisfying is $8$

Using $\large ^{n-1}C_r + ^{n-1}C_{r-1} = ^nC_r \ :$ $\binom{n-1}{3} + \binom{n-1}{4} = \binom{n}{4}$ Using pascal's triangle it is clear to see that $\binom{2n}{n} > \binom{2n}{n \pm1} > \binom{2n}{n \pm 2} >...> 1$ $\therefore Min(n) = 8$

$\begin{aligned} \color{#3D99F6} {n-1 \choose 3} + {n-1 \choose 4} & > {n \choose 3} & \small \color{#3D99F6} \text{Pascal's formula: } {n \choose r} = {n-1 \choose r} + {n-1 \choose r-1} \\ \color{#3D99F6} {n \choose 4} & > {n \choose 3} \\ \frac {n!}{4!(n-4)!} & > \frac {n!}{3!(n-3)!} \\ n-3 & > 4 \\ n > 7 \\ \implies \min (n) & = \boxed{8} \end{aligned}$