+54 An organization has 50 people, and 4 of the people are running for president. Each of the 50 people votes for one of the 4 candidates. How many different vote totals are possible?
+2095 Here is a hint to make everything clearer:
Let's say that the people voting are Peoples 1-50.
The candidates are Candidates a-d.
We can draw a diagram to represent this question, or we can find the equation easily, labeling x as the people voting and y as the candidates.
Hope this helped!
+499 https://web2.0calc.com/questions/can-t-solve-this-blame-skyrim
This question has been posted maybe 6 times already in the past day - 2. I solved part of it here
+54 I'm confused on what equation is needed to solve this problem. I don't know which formula or something that relates to this problem. Also, I have seen the second part, but all I need to solve is the first part.
+499 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
For stars and bars:
like the above example, just think of the votes as 50 "balls" that are divided by 3 dividers, which create 4 regions for vote allocation(for 4 candidates). the formula for stars and bars(non negative, meaning 0 is allowed) is:
(n + k -1) choose (n)
where n is the number of things you want to distribute or "stars", and k is the number of groups you want to distribute those stars into.
+499 This is a really common method of counting known as "stars and bars". If you want more reference information, I'd highly recommend looking to wikipedia or other sources online. The basic idea is that we can think of 50 votes as "50 balls" or "stars" laid out in a row.
These 50 stars are divided by 3 "bars", which form 3 groups. You can picture it like this:
if the O's are balls, and the |'s are dividers, we have:
O O O O O O | O O O O O | O O O O | O O .. ..... x 50.
These bars divide how many votes each of the four candidate gets(can you see why there are 3 bars? because if n is the number of bars there are, n+1 regions are formed from n bars).
Because we want to see how many votes each candidate could get, the answer is stars and bars' non negative form(there are two forms of stars and bars, watch out!)
With 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
+499 ok I'm not really sure why, but web2.0 calc keeps constantly flagging my responses for this question for moderation. I'll just explain the basic jist of it. This problem employs a common counting method known as "stars and bars." with stars and bars, you can find out how many ways there are to distribute n things into k groups. For stars and bars, there are two forms of it: non negative and positive stars and bars(in other words, non negative is where a group is allowed to have 0 things in it, and positive at least 1 thing). This problem is asking for the non negative form of stars and bars, which is :
\({n+k-1 \choose n}\) With this formula, the problem becomes a lot easier!
+54 Wow! Thank you guys so much for helping me! I learned so much from you guys! Hope you guys stay safe and have a good day!
+118673 Hi all (on this thread)
I have just unflagged all jfan17 's posts. So you can see them all properly now. 
+118673 Thanks jfan17,
I would do it the same way but I rarely bother with any but the most basic formulas
I just think of it like this.
There are 50 votes. (the stars)
We need to divide them into 4 groups which will need 3 bars .
The first candidate will get all the votes in front of the first bar.
The second candidate gets all the votes between the 1'st and the 2'nd bar.
The third candidate gets all the votes between the 2'st and the 3'nd bar.
The fourth candidate gets all the votes after the 3rd bar.
So there are 53 things. 50 votes and 3 bars.
All we need to do ist to work out how many different combinations are possible for where to put the divider bars
So it will be 53C3 ( which is exactly the same as 53C50 )
