How many election outcomes in the race for class president are there if there are five candidates and 40 students in the class and exactly three candidates tie for the most votes?

We have 40 students, five candidates, and the top three tie for the most votes. How many different election results can we achieve?

If we only care about the number of ways we can see who won and not about the vote count, we can see that there is a pool of five, three of whom will win. So that gives us:

#C_(5,3)=(5!)/((3)!(5-3)!)=(5!)/((3!)(2!))=(5xx4xx3!)/(3!xx2)=10#

where

#C_(n,k)=(n!)/((k)!(n-k)!)# with #n="population", k="picks"#

If we do care about the vote count, read on...

We know that there are 40 votes cast and there is a three-way tie for the most votes. I'm going to call the vote count for any one of the top three candidates #M# (for the Most), and the other two can be #x# and #y#.

We have to multiply M by 3, and so we'll have the vote count be:

#3M+x+y=40; M>x, y#

Let's first work with #M#

How large can #M# be? That would be the case where the top three candidates got almost, if not all, the votes:

#40/3=13 1/3=> color(blue)(ul(bar(abs(color(black)M<=13))))#

How small can #M# be? Let's first see that:

#40/5=8# - if the votes had been cast evenly across all five candidates, each of them would have received 8 votes. So what would happen if only the top three candidates got 8 votes? Then they'd have:

#3xx8=24# votes, with 16 left over. There is no way to split 16 votes between #x# and #y# and meet our conditions:

• #x, y !=8# because that would then be a 5-way tie
• which means either #x# or #y# would have to be greater than 8, which would mean the 3 candidates who got 8 votes wouldn't have a tie for the most votes.

And so #color(blue)(ul(bar(abs(color(black)(M>=9))))#

Ok - we know that #9 <= M <= 13#, or 5 different vote counts (9, 10, 11, 12, 13). Which means that #x, y# need to soak up the remaining votes.

If we don't care about the details of the results for the two losing candidates, we can see that for any value of #M#, there is a pool of five people that can be in any of three slots, and so that's the combination:

#C_(5,3)=(5!)/((3)!(5-3)!)=(5!)/((3!)(2!))=(5xx4xx3!)/(3!xx2)=10#

where

#C_(n,k)=(n!)/((k)!(n-k)!)# with #n="population", k="picks"#

There are five different values for #M#, and so there are:

#5xx10=50# election results.

~~~~~

What if we do care about the votes received by the losing candidates?

How many votes can #x,y# have?

There are a few limitations:

• We know that #x,y < M#
• #x,y >=0#
• #x+y=40-3M#

Let's try a table to summarize allowable values of #x# with different values of #M#:

#((M,3M,"allowable x","resulting y","count of allowable x, y"),(13,39, "0, 1","1, 0",2),(12,36,"0, 1, 2, 3,4","4, 3, 2, 1, 0",5),(11,33,"0, 1, 2, 3, 4, 5, 6, 7","7, 6, 5, 4, 3, 2, 1, 0",8),(10, 30, "1, 2, 3, 4, 5, 6, 7, 8, 9","9, 8, 7, 6, 5, 4, 3, 2, 1",9),(9, 27, "5, 6, 7, 8", "8, 7, 6, 5",4))#

Now we know how many ways the votes can be cast. Now we need to put candidates to the vote count.

We know that #C_(5,3)=10# and the remaining vote count #x# can be held by one of the other two people, forcing the #y# value to be held by the last candidate. This means that for each vote count, there are #10xx2=20# different ways it can be achieved.

There are a total of #2+5+8+9+4=28# different ways the vote count can turn out, and so there are:

#20xx28=560# ways the election outcomes can happen.