I know that there is a formula where even 0 is treated as significant, but is there a formula where 1 and 0 cannot be parts?


Edit:The formula that I was referencing was (Excuse the presentation) $\frac{(n-1)!}{(k-1)!(n-k)!}$
This one discards the 0, but what I was looking for one that would discard 1 as well. 
Also, what I'm looking for is the compositions, not the partitions. 



Number of compositions of n into k parts (where 0 and 1 are not allowed as parts)
Here how to obtain the generating function:


Give a compsotion of $n$ with $k$ summands the weight $x^n y^k$. The compositions of $n$ into $k$ parts with terms $\geq 2$ are then obtained as $Sequ(\mathbb{N}\_{\geq 2})$, see for example 2.2.1. of <https://arxiv.org/pdf/1409.2562.pdf>.
Thus the generating function is $\frac{1}{1-(x^2y+x^3y+....)}=\frac{1-x}{(1-x)-yx^2}$.


Now you can get a formula by expanding the function $\frac{1-x}{(1-x)-yx^2}$.


$\frac{1-x}{(1-(x+yx^2)}=(1-x) \sum\limits\_{k=0}^{\infty}{(x+yx^2)^k}=(1-x) \sum\limits\_{k=0}^{\infty}(x^k \sum\limits\_{l=0}^{k} \binom{k}{l} (yx)^l)$.
From this one should be able to order the terms together for the form $y^s x^t$ to arrive at the result $\binom{n-k-1}{k-1}$ as in the comment.


