Let $\mathcal{F}\subseteq 2^{[m]}$ be a family of subsets of $[m]=\{1,2,\ldots,m\}$. For $S\subseteq [m]$, let $\mathcal{F}|_S$ be the trace $\mathcal{F}|_S=\{B\cap S : B\in\mathcal{F}$, considered as a multiset. We say $\mathcal{F}$ shatters a set $S\subseteq [m]$ if $\mathcal{F}|_S$ has all $2^{|S|}$ possible sets (i.e. complete). We say $\mathcal{F}$ has a shattered set of size $k$ if $\mathcal{F}$ shatters some $S\subseteq [m]$ with $|S|=k$. It is well known that if $\mathcal{F}$ has no shattered $k$-set then $|\mathcal{F}|\le\binom{m}{k-1}+\binom{m}{k-2}+\cdots+\binom{m}{0}$. We obtain the same exact bound when forbidding less. Namely, given fixed positive integers $t$ and $k$, for every set $S\subseteq [m]$ with $|S|=k$, $\mathcal{F}$ is such that $\mathcal{F}|_S$ does not have both all possible sets $2^S$ and specified additional sets occuring at least $t$. Similar results are proven for double shattering, namely when $\mathcal{F}|_S$ does not have all sets $2^|S|$ appearing twice. The paper is written in matrix notation with trace replaced by configuration.