Another Limit to Limit

Similar solution with @Aaghaz Mahajan 's

Consider the inequality

$\begin{aligned} \binom m{n-1} & > \binom {m-1}n \\ \frac {m!}{(n-1)!(m-n+1)!} & > \frac {(m-1)!}{n!(m-n-1)!} \\ \frac m{(m-n)(m-n+1)} & > \frac 1n \\ mn & > m^2 - 2nm + n^2 + m - n & \small \color{#3D99F6} \text{Rearranging} \\ m^3 - (3n-1)m + (n^2-n) & < 0 \\ \implies m & < \frac {3n-1+\sqrt{(3n-1)^2-4(n^2-n)}}2 \\ M(n) & < \frac {3n-1+\sqrt{5n^2-2n+1}}2 \\ \implies \lim_{n \to \infty} \frac {M(n)}n & = \lim_{n \to \infty} \frac {3n-1+\sqrt{5n^2-2n+1}}{2n} = \frac {3+\sqrt 5}2 \end{aligned}$

Therefore $a+b+c = 3+5+2 = \boxed{10}$ .

We need

$\frac{m!}{\left(n-1\right)!\cdot\left(m-n+1\right)!}>\frac{\left(m-1\right)!}{n!\cdot\left(m-n-1\right)!}$

After simplifying, we need to solve for m in the quadratic inequality

$m^2-\left(3n-1\right)m+\left(n^2-n\right)<0$

For, the largest value of m for a given n, we simply need the bigger root of the inequality given.....i.e.

$M\left(n\right)=\frac{3n-1+\sqrt{5n^2-2n+1}}{2}$

Now, simply taking the limit of the desired expression, we get the value of the limit, $\displaystyle \boxed{\frac{3+\sqrt{5}}{2}}$