A Confused Equation!

By the AM-GM Inequality, we have:

$\dfrac{\binom{n}{n-1}^6 + \binom{n-2}{k}^6 + \binom{n+3}{n+1}^3 }{3 }$ $\geq \sqrt[3]{\binom{n}{n-1}^6 \binom{n-2}{k}^6 \binom{n+3}{n+1}^3} = \binom{n}{n-1}^2 \binom{n-2}{k}^2 \binom{n+3}{n+1} = n^2 \binom{n-2}{k}^2 \binom{n+3}{2}$

where equality holds if and only if:

$\binom{n}{n-1}^6 = \binom{n-2}{k}^6 = \binom{n+3}{n+1}^3$

which is equivalent to:

$n^2 = \binom{n-2}{k}^2 = \binom{n+3}{2}$

Comparing the inequality obtained to the original equation, we see that we need the equality in our inequality in order for the original equation to be true.

Solving $n^2 = \binom{n-2}{k}^2 = \binom{n+3}{2}$ , we get $n=6$ and $k=2$ only.

Thus $n+k = 6+2 = \boxed{8}$