Find a closed formula for the stirling number $S(n, n-3)$ for n $\ge$ $3$

Question

I'm trying to find the Stirling number $S(n, n-3)$ for $n$ $\ge$ $0$.

(the number of ways to place $n$ distinct objects, into $n-3$ identical boxes)

Obviously the first option is choosing the first box to contain a set of $4$ elements while the rest are singletons.

Giving : ${n \choose4}$

But after this I'm stuck.

The other options of picking are:

1. 3 elements in the first box, 2 in the second while the rest are singletons
2. 2 elements in the first box, 2 in the second and, 2 in the third, rest singletons.

For the second one, my best guess would be $\frac{1}{6}{n\choose2}{n-2\choose2}{n-4\choose2}$.

Any help on 1. and 2.?

Answer 1

Combining the three possible outcomes of choices we get :

$$S(n,n-3)={n\choose 4} + \frac{1}{6}{n\choose 2}{n-2\choose 2}{n-4\choose 2} + \frac{1}{2}{n\choose 3}{n-3\choose 2}$$

Answer 2

You can verify my answer with the table of S(n,k) values on Wikipedia. It works.

Not sure if an answer without discrete cases (like n = 3, 4, 5) can be derived, but at very least this answer is absolutely correct..