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#1
11-07-2019, 08:38 PM
 hostess Member CAS Join Date: Aug 2018 Posts: 820
Combinatorics Voting Problem

In approval voting, each voter can distribute up to 5 votes among 6 candidates. For example, you could cast 3 votes for one candidate and 2 for another, or you could cast 1 vote for each of 4 candidates (and not cast your fifth vote). In how many ways can you distribute your votes?

My thought was: you have 5 votes and 7 choices for each (6 candidates + no vote)... so do 7^5... and order doesn't matter so divide by 5!. This results in a decimal.

What am I doing wrong here
#2
11-07-2019, 08:46 PM
 Colymbosathon ecplecticos Member Join Date: Dec 2003 Posts: 6,167

Quote:
 Originally Posted by hostess In approval voting, each voter can distribute up to 5 votes among 6 candidates. For example, you could cast 3 votes for one candidate and 2 for another, or you could cast 1 vote for each of 4 candidates (and not cast your fifth vote). In how many ways can you distribute your votes? My thought was: you have 5 votes and 7 choices for each (6 candidates + no vote)... so do 7^5... and order doesn't matter so divide by 5!. This results in a decimal. What am I doing wrong here
Suppose you want to cast just one vote for candidate A. You can do this 5 ways, not 120 ways.
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#3
11-07-2019, 09:07 PM
 hostess Member CAS Join Date: Aug 2018 Posts: 820

Quote:
 Originally Posted by Colymbosathon ecplecticos Suppose you want to cast just one vote for candidate A. You can do this 5 ways, not 120 ways.
ok so I guess you only divide by the factorial of how many unique candidates you are voting for... so my answer would change to this

Situation where you vote for the same candidate (including none) 5 times: You have 5 votes and 7 choices which all have to be the same.... thus you have 7 ways to do this

Situation where you vote for 2 different candidates (including none): You have 5 choices and 7 candidates... The first one can be anyone, the second can be any of the remaining 6 candidates, and the other 3 votes have to be between the first two candidates you picked... order doesnt matter so we get: 7*6*(2^3)/2! ways to do this

situation where you vote for 3 candidates... 7 choices for first, 6 for second, 5 for third, and you must choose between first 3 candidates for the last 2 votes... order does not matter, so 7*6*5*3^2/3! for this...

continue this to 5 unique candidates situation and you sum up all the components resulting in:

7+
7*6*(2^3)/2! +
7*6*5*(3^2)/3! +
7*6*5*4*(4^1)/4! +
7*6*5*4*3*(5^0)/5!

which equals 651... not the correct answer
#4
11-07-2019, 09:25 PM
 Colymbosathon ecplecticos Member Join Date: Dec 2003 Posts: 6,167

How many ways can you cast exactly 0 votes?
How many ways can you cast exactly 1 vote?
How many wats can you cast exactly 2 votes?

How many ways can you cast exactly 5 votes?

The sum of those is the total number of ways.

This is a hard way, but usually there is an way to do the counting. Try counting something else.
__________________
"What do you mean I don't have the prerequisites for this class? I've failed it twice before!"

"I think that probably clarifies things pretty good by itself."

"I understand health care now especially very well."
#5
11-07-2019, 10:09 PM
 Colymbosathon ecplecticos Member Join Date: Dec 2003 Posts: 6,167

Hint: You're looking at votes and asking how can these be cast.

Try looking at candidates and asking how many votes did each get.

Hint: not voting is a vote for OTHER
__________________
"What do you mean I don't have the prerequisites for this class? I've failed it twice before!"

"I think that probably clarifies things pretty good by itself."

"I understand health care now especially very well."
#6
11-08-2019, 01:37 PM
 hostess Member CAS Join Date: Aug 2018 Posts: 820

Quote:
 Originally Posted by Colymbosathon ecplecticos How many ways can you cast exactly 0 votes? How many ways can you cast exactly 1 vote? How many wats can you cast exactly 2 votes? … How many ways can you cast exactly 5 votes? The sum of those is the total number of ways. This is a hard way, but usually there is an way to do the counting. Try counting something else.
Yeah I know the combinations with repetition formula, but I'm trying to solve it the hard way to explain it to another student.
lets say we have candidates a,b,c,d,e,f
cast exactly 0 votes: 6^0 = 1 way
cast exactly 1 vote: 6^1 = 6 way
cast exactly 2 votes: AA, AB, AC, AD, AE, AF, BB, BC, BD, BE, BF, CC, CD, CE, CF, DD, DE, DF, EE, EF, FF which is 21 ways ... now how can i formalize this? if you chose the same candidate twice you have 6 options. if you choose different candidates, you have 6*5/2 options... so 15+6=21
cast exactly 3 votes: 6 choices if you chose the same candidate thrice.

if you choose 2 unique candidates, you have 6*5/2 = 15 choices.

if you choose 3 unique candidates you have 6*5*4/3! = 20 choices... add em all up and you get 41 choices..

cast exactly 4 votes: 6+15+20+6*5*4*3/(4!) =56 choices

cast exactly 5 votes: 6+15+20+15 + 6*5*4*3*2/(5!) =62 choices

Add them up and you get 62+56+41+21+6+1=187? Something seems wrong here.. jesus i feel dumb... i'll take a look at this after work
#7
11-08-2019, 04:11 PM
 Colymbosathon ecplecticos Member Join Date: Dec 2003 Posts: 6,167

You've made the problem much harder than it is.

You have 5 votes to cast among 7 candidates, the 7th candidate being OTHER as you observed.

Let Ni be the number of votes you cast for candidate i.

Your vote constraint is that the sum if the Ni (i=1,...7) = 5 (all uncast votes go to OTHER).

The other constaints are that each Ni is an integer and is >=0.

So the question is equivalent to: "How many non-negative, integer (vector) solutions are there to sum (i=1,...,7)Ni = 5?"

Most probability books show how to solve that counting problem.
__________________
"What do you mean I don't have the prerequisites for this class? I've failed it twice before!"

"I think that probably clarifies things pretty good by itself."

"I understand health care now especially very well."