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#1




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




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#3




Quote:
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




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




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




Quote:
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




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 nonnegative, 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." 
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