Every entity must receive at least 2 cards and all cards have to be distributed. Only the cards and not the entities are distinguishable. My approach was as follows:
Since every entity has to receive at least 2 cards, I first pick $n$ cards, i.e. $m\choose n$. The remaining $m-n$ cards can then be distributed across the $n$ entities using Stirling numbers of the second kind, guaranteeing that each entity receives its remaining card and that all remaining cards are distributed without distinguishing the entities. The final construct is then:
$${m\choose n}\cdot S(m-n, n)$$
The actual solution, however, is: $$\frac{17!}{6!1!(2!)^6(5!)^1}+\frac{17!}{5!1!1!(2!)^5(3!)^1(4!)^1}+\frac{17!}{4!3!(2!)^4(3!)^3}$$ which uses the fact that the number of equivalence relations over $[N]$ which contain exactly $\lambda_i$ many $i$-element equivalence classes is given by: $$\frac{N!}{\lambda_1!\lambda_2!..\lambda_N!(1!)^{\lambda_1}(2!)^{\lambda_2}..(N!)^{\lambda_N}}$$ and that given the requirements of the task, the 17 cards can be distributed across the 7 entities in three ways: (2, 2, 2, 2, 2, 2, 5) (i.e. six $2$-element classes and one $5$-element class), (2, 2, 2, 2, 2, 3, 4), or (2, 2, 2, 2, 3, 3, 3).
This gives a much larger number than my solution. In what way do the two differ and what is my formulation actually describing?