My question is most easily explained via an example: Consider the number $n=1967$. It contains the prime numbers $7$, $19$, $67$ and $967$ as all those are primes and you can find them in the decimal expansion of $1967$. In contrast, the part $196$ for example is not prime.
Given a fixed number of digits, $k$ say, what can be said about the maximal number of primes you can find in the decimal expansion of a $k$-digit number? I will call this number $m(k)$. The above example then teaches us that $m(4)\ge 4$.
Clearly, $m(k) \ge k$ e.g. by considering the number with digit $3$ $k$-times. The same argument shows that $m(k+1)\ge m(k)+1$. The maximal number of subnumbers for a $k$-digit number is $\frac{k(k+1)}{2}$, so trivially $m(k)\le \frac{k(k+1)}{2}$.
The first cases are: $$m(1)=1\ (e.g. n=3)\\ m(2)=3\ (e.g. n=23)\\ m(3)=6\ (e.g. n=373).$$