### Instruction: At the end of the year, the Math Club decided to hold an election for which 5 equal officer positions were available. However, 16 candidates were nominated, of whom 7 were past officers. Of all possible elections of the officers, how many will have at least 1 of the past officers?### Response: I think we should use complementary counting.
That means we should find the number of elections that have no past officers and subtract that from the total number of possible elections.
That's correct. First, let's find the total number of possible elections.
Each candidate can either be elected or not elected, so there are 2 options for each candidate.
That's right. Since there are 16 candidates, there are $2^{16}=65536$ possible elections.
Now we need to find the number of elections that have no past officers.
That means we need to find the number of possible elections using only the 9 non-past officers.
Each of these 9 candidates can either be elected or not elected, so there are 2 options for each candidate.
So there are $2^9=512$ possible elections that have no past officers.