Difference between distributing (distinguishable/indistinguishable) objects into groups

Question

I'm having a bit of problem doing the following:

1) Number of ways to distribute 7 (different) students in 4 (non empty) teams:

1. $490$
2. $350$
3. $560$
4. $210$

My understanding this is the Stirling number of the second kind and we can use inclusion-exclusion: $$S(7,4)=\frac{1}{4!}\sum_{j=1}^4(-1)^{4-j}\binom{k}{j}j^n=\frac{1}{24}(-4+768-8748+16384)=\frac{8400}{24}=350.$$

2) Number of ways to distribute 7 tutors into 4 departments where each department must get at least 1 tutor:

1. $350$
2. $4200$
3. $1400$
4. $8400$

I don't quite see the difference here. How does the fact that the "tutors" are not distinguishable affect this?

Answer 1

People are considered to be distinguishable unless otherwise specified. The difference between the two questions is that departments are distinguishable while unlabeled teams are not.

Number of ways to distribute $7$ students to $4$ (nonempty) teams.

Your answer of $S(7, 4) = 350$ is correct since the teams are not labeled.

Number of ways to assign $7$ tutors to $4$ departments so that each department receives at least one tutor.

Since departments are distinguishable (the mathematics department is different from the physics department), the answer to this question is $$4!S(7, 4) = \sum_{k = 0}^{4} (-1)^k \binom{4}{k}(4 - k)^7 = 4^7 - \binom{4}{1}3^7 + \binom{4}{2}2^7 - \binom{4}{3}1^7 + \binom{4}{4}0^7 = 8400$$ where $\binom{4}{k}$ is the number of ways of excluding $k$ of the $4$ departments from receiving a tutor and $(4 - k)^7$ is the number of ways of distributing the $7$ students to the remaining $4 - k$ departments.