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An organization has 50 people, and 4 of the people are running for president.

 

Part A) Each of the 50 people votes for one of the 4 candidates. How many different vote totals are possible?

Part B) Each of the 50 people either votes for one of the 4 candidates, or can abstain from voting. How many different vote totals are possible?

 Mar 20, 2020
 #1
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I've tried posting answers to this before but the explanations were too long so I'll post it in bits:

 

1.  This is a textbook case of a method of counting known colloquially as "stars and bars" or "balls and urns"(I'd recommend you to search it up online). We can think of the 50 votes as 50 balls laid out in a row, divided by 3 "dividers" which separate the balls into 4 separate regions for each candidate, with each region being how many votes a candidate gets. Using stars and bars, we get the equation:

(50+4-1) c (50),  or (53c50). This gives us 53*52*51/3!

 

This simplifies to:

 

53*26*17 = 23426 totals

 Mar 21, 2020
 #2
avatar+402 
+4

2. So as for Part b, it becomes a little trickier. Imagine the problem as the same as the first part, except there are 4 dividers. You might be wondering why there are four dividers, since that produces 5 "regions", but there are only 4 candidates! However, you need the 5th "region" to count for people who abstain(which is a clever trick you can use for problems like this) from voting for a candidate; you could almost think of the people who abstain as people who vote for another candidate. With this knowledge in hand, the problem becomes a lot like the first problem, and with stars and bars, we get:

 

(50+5-1) c (50) or (54c50). This gives us 54*53*52*51/4!, or 316251 totals

 

EDIT: Since this post finally got through, my answer to part A was flagged for moderation, but using stars and bars, we can find the answer to both the first and second parts. However, Part B just puts a "spin" on the stars and bars technique as I mention above

 Mar 21, 2020
edited by jfan17  Mar 21, 2020
 #3
avatar+402 
+2

1.  This is a textbook case of a method of counting known colloquially as "stars and bars" or "balls and urns"(I'd recommend you to search it up online). We can think of the 50 votes as 50 balls or stars laid out in a row, divided by 3 "dividers" or bars which separate the balls into 4 separate regions for each candidate, with each region being how many votes a candidate gets. With this, we can think of how many possibles "spaces" you can choose for the dividers to go in. Using stars and bars, we get the equation:

(50+4-1) c (50),  or (53c50). This gives us 53*52*51/3!

 

This simplifies to:

 

53*26*17 = 23426 totals

 Mar 21, 2020
 #4
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+1

great! Thank you so much!

Guest Mar 23, 2020

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