There is 15 voters, and 4 candidates. The candidates are part of these 15 voters, and they can vote on themselves. Each voter must cast a vote. Now, how many possible ways can the votes be cast if at least one candidate receives no votes?
My idea for a solution would be to add up cases where at least one candidate is not a voting option. Using $\binom{m+n-1}{n}$ where m is candidates and n is voters, I would add $\binom{3+15-1}{15}$ and $\binom{2+15-1}{15}$ and $\binom{1+15-1}{15}$. I think that is a valid way of approaching the problem with combinatorics.
However there is a hint below that reads "Let $A_i$ be the set of all outcomes where person $i$ receives no votes. Now use principle of inclusion exclusion.
How would I write that set $A_i$? And then how would I use PIE on said set? This way of approaching the problem seems complicated and I am unsure of where to start. Even a nudge in the right direction would be appreciated.
You want to count $|A_1\cup A_2\cup A_3\cup A_4|$. By inclusion exclusion, this is $$ \sum_i |A_i|-\sum_{i<j}|A_i\cap A_j|+\sum_{i<j<k}|A_i\cap A_j\cap A_k|-|A_1\cap A_2\cap A_3\cap A_4|. $$ Now, $$ \begin{align} |A_i|&=\binom{15+3-1}{15}\\ |A_i\cap A_j|&=\binom{15+2-1}{15}\\ |A_i\cap A_j\cap A_k|&=\binom{15+1-1}{15}\\ |A_1\cap A_2\cap A_3\cap A_4|&=\binom{15+0-1}{15}=0\\ \end{align} $$ Finally, counting the number of times each term is repeated, you get $$ |A_1\cup A_2\cup A_3\cup A_4|=\binom41\binom{15+3-1}{15}-\binom42\binom{15+2-1}{15}+\binom43\binom{15+1-1}{15} $$