The question essentially states:
The seven dwarfs decide to form four teams for a competition. The teams can be of varying sizes, but there must be at least one dwarf in a team, and a dwarf can only be part of one team. The order of the teams or of the dwarfs within
the teams does not matter. In how many ways can the four teams be made up?
I considered how many partitions of $7$ into $4$ groups exist. Ie
Case A: $1,1,1,4\\$
Case B: $1,1,2,3\\$
Case C: $1,2,2,2\\$
For case A we have $7C4$ possibilities
For case B we have $7C3 \times 4C2$
For case C we have $7C2 \times 5C2 \times 3C2$
Is this approach correct? If not how could it be amended to get the right answer? What are some other ways of thinking about this problem?
Best Answer
Your approach is essentially correct. As pointed out by Lord Shark the Unknown in a comment, your calculation for $C$ isn't quite right because you calculated as if the order in which you choose the teams of $2$ mattered; you need to divide by $3!$ because it doesn't. Alternatively, you can directly use the multinomial coefficient $\binom7{2,2,2}$. Then your approach yields
$$ \binom74+\binom73\binom42+\frac1{3!}\binom7{2,2,2}=35+35\cdot6+\frac16\cdot630=350 $$
in agreement with Hat's answer. As for other approaches, the Stirling numbers have already been pointed out. If you don't want to learn about a specific type of numbers and would rather learn a more general method, you could apply the inclusion–exclusion principle. The expression that Hat used for the Stirling numbers is also proved by applying that principle, so in a sense the Stirling numbers are just a shorthand for the inclusion–exclusion principle.