Combinatorics involving voting A disciplinary hearing has seven panelists who wish to vote on whether a member of an organization should be expelled from the organization or not. Each panelist can either vote for or against the expulsion of the member or abstain from voting. The member can only be expelled if at least 5 panelists vote for his expulsion. In how many ways can the outcome of the voting process result in the member being expelled? The answer is 99 but I have no idea as to how to get there. Please help.

Let's represent a vote FOR expulsion as $0$, AGAINST expulsion as $1$, and ABSTAIN as $2$. Each committee member is distinguishable, so we may view the votes as a ternary string of length $7$. The goal is to count the number of ternary strings with at least five $0$'s.

For $i \in \\{5, 6, 7\\}$, we select the $i$ positions for $0$'s in $\binom{7}{i}$ ways. The remaining $7-i$ positions can have either $1$'s or $2$'s. So by rule of product, there are $\binom{7}{i}2^{7-i}$ ternary strings of length $7$ with $i$ $0$'s.

By the rule of sum, we add up from $i = 5$ to $i = 7$:

$$\sum_{i=5}^{7} \binom{7}{i}2^{7-i}$$