# How many election outcomes in the race for class president are there if there are five candidates and 40 students in the class and no candidate receives a majority of the votes, meaning nobody gets higher than 20 votes?

See below on a discussion of a possible methodology:

#### Explanation:

I don't know of a simple way to solve this, but if I was to tackle this problem, this is the methodology I'd use:

• let's first see that there are going to be 40 votes total and 5 candidates, and so we'll be looking at how to distribute those 40 votes across the 5 candidates,
• let's next see that the problem of having duplicate vote tallies will need to be accounted for,
• and so to accomplish this, we'll need to list out the possible vote results across First Place, Second Place, etc, and then assign candidates to those places using permutations and then divide by an appropriate factor to eliminate the duplicates.

Vote Tallies

This is where the majority of the work lies. Unfortunately, I don't see any way to save time or energy on this part. Each unique vote tally needs to be listed.

We know that there are 40 votes and no one got above 20, and so one vote tally can be (20, 20, 0, 0, 0). The next vote tally can be (20, 19, 1, 0, 0). And so on (each following place will have less than or equal to the place before). We'll work our way down this list until we get to (8, 8, 8, 8, 8).

$\left(\begin{matrix}\text{First" & "Second" & "Third" & "Fourth" & "Fifth} \\ 20 & 20 & 0 & 0 & 0 \\ 20 & 19 & 1 & 0 & 0 \\ 20 & 18 & 2 & 0 & 0 \\ 20 & 18 & 1 & 1 & 0 \\ 20 & 17 & 3 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 20 & 5 & 5 & 5 & 5 \\ 19 & 19 & 2 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 8 & 8 & 8 & 8 & 8\end{matrix}\right)$

Permutations

We can now assign candidates to each place. There are 5! =120 ways to do that for each unique vote tally. However, we divide by the factorial of the number of duplicate vote counts to eliminate duplicates. For instance, with the first vote tally of (20, 20, 0, 0, 0), there are two vote counts that have duplicates - there are three 0s and two 20s, so we divide the 5! by 3!2!, which gives us:

(5!)/(3!2!)=120/(6xx2)=10 ways in which this result can occur.

For the next unique vote tally of (20, 19, 1, 0, 0), there are two 0s, and so there are:

(5!)/(2!)=120/2=60 ways in which this can occur.

And we do this for each unique vote tally, all the way to (8, 8, 8, 8, 8) which has only 1 way it can occur ((5!)/(5!)=1)

It is the number of ways that each vote tally can occur that will be summed up and give the final result.

(("First", "Second", "Third", "Fourth", "Fifth",5!,-:,"Ways"),(20,20,0,0,0,120,12,10),(20,19,1,0,0,120,2,60),(vdots, vdots,vdots,vdots,vdots,120,vdots,vdots),(20,5,5,5,5,120,24,5),(19,19,2,0,0,120,4,30),(vdots, vdots, vdots, vdots, vdots,120,vdots, vdots),(8,8,8,8,8,120,120,1))