Find the smallest value of $n$ such that row $n$ of Pascal's triangle contains three successive entries with the ratio $3:4:5$.
I was able to solve this with algebraic manipulation. We have$${{\binom{n}{k}}\over3} = {{\binom{n}{k+1}}\over4} = {{\binom{n}{k+2}}\over5}$$This can be written as$${{n!}\over{3(n-k)!k!}} = {{n!}\over{4(n-k-1)!(k+1)!}}, \quad {{n!}\over{4(n-k-1)!(k+1)!}} = {{n!}\over{5(n-k-2)!(k+2)!}}$$This can be manipulated to$$4(k + 1) = 3(n - k), \quad 5(k+2) = 4(n - k - 1)$$Simplifying, when we solve for $k$ in both equations:$$k = {{3n - 4}\over7} = {{4n - 14}\over9}$$Now solving for $n$, we get $n = 62$.
However, I am wondering if there a direct (as possible) combinatorial way to obtain this result as opposed to the algebraic way I demonstrated here.