A person works at 1 to 3 of 5 different places and eats 1 to 3 of 4 different meals every day. Prove there were two days in a year where he ate the exact same meals and worked at the exact same place.
MY ANSWER
The ensemble of combinations can be described as the sum of the sets of workplaces: $\sum_{i=1}^{3}\operatorname{nCr}\left(5,i\right)$ times the sum of sets of meals: $\sum_{n=1}^{3}\operatorname{nCr}\left(4,n\right)$. $$\sum_{n=1}^{3}\operatorname{nCr}\left(4,n\right)\cdot\sum_{i=1}^{3}\operatorname{nCr}\left(5,i\right)=350$$ Since $350<365$ then the 15 last days must be repetitions of a combination that happened in the initial 350 days.
Is this a proper reading of the situation?
The math is fine, but $\mathrm{nCr}(4,n)$ is bizarre. In ${}_nC_r$ notation for binomial coefficients, you would just write ${}_4C_n$. But it's more common to write $C^4_n$, and it's even more common than that to write $\binom4n$. So I would write $$ \left(\binom51 + \binom52 + \binom 53\right) \cdot \left(\binom 41 + \binom42 + \binom43\right) $$ which simplifies to $(5 + 10 + 10) \cdot (4 + 6 + 4) = 25 \cdot 14 = 350$. The summation notation $$\sum_{i=1}^3 \binom5i \cdot \sum_{j=1}^3 \binom 4j$$ is also fine.
Additionally, getting $350 < 365$ tells us that there must be at least $15$ days that are repeats of previous days (which is all you need to know here), but it does not tell us that the last $15$ days are the repeats. For example, maybe the person worked at the same place and ate the same thing for the first $100$ days of the year, and then decided to keep trying new options for the next $265$ days.