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Problem

Suppose, we have a voting with n voters and m candidates. Each voter can vote for 1 candidate. How many possible outcomes of voting can be?

Solution I tried

For 1 candidate there will always be 1 outcome.

Possible outcomes for 2 candidates:

1 voter        2 voters       3 voters       4 voters
 |1|0|         |2|1|0|        |3|2|1|0|     |4|3|2|1|0|
 |0|1|         |0|1|2|        |0|1|2|3|     |0|1|2|3|4|
2 outcomes    3 outcomes     4 outcomes     5 outcomes

So, we have a sequence: 2, 3, 4, 5, ..., n + 1, ...

3 candidates

 1 voter          2 voters                3 voters
 |1|0|0|       |2|0|0|1|1|0|      |3|0|0|2|2|0|0|1|0|1|
 |0|1|0|       |0|2|0|1|0|1|      |0|3|0|1|0|2|1|0|1|1|
 |0|0|1|       |0|0|2|0|1|1|      |0|0|3|0|1|1|2|2|2|1|
3 outcomes      6 outcomes             10 outcomes

3,6,10, ... ?

As far as I understand, first we need to take total number of groups of n elements with m elements in group: $n^m$

For example in case of 1 voter and 3 candidates total will be $2^3 = 8$:

|0|0|1| <- good
|0|1|0| <- good
|1|0|0| <- good
|1|0|1|
|1|1|0|
|1|1|1|

These are possible binary numbers, consisting of 3 positions.

And find the number of those numbers, whose sum of digits is equal to n. But I don't know, how to do the last part.

user4035
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3 Answers3

5

Hint: We will first assume (dangerous assumption) that every voter votes. Then this is a standard "Stars and Bars" problem. The Wikipedia article is quite good.

We want to find the number of solutions in non-negative integers of the equation $x_1+x_2+\cdots +x_m=n$.

You will see by looking at the article that the answer can be written as $\dbinom{m+n-1}{n}$.

If we assume that perhaps not every voter votes, make up a dummy candidate, that we can call Candidate Nobody. We can say that the people who didn't vote voted for Candidate Nobody.

We are then finding the number of non-negative integer solutions of $x_0+x_1+\cdots+x_m=n$.

André Nicolas
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  • Checking my manual solutions with your formula now – user4035 Sep 01 '13 at 17:56
  • My solutions were correct, will read, how this formula was deduced now. Thanks :). – user4035 Sep 01 '13 at 18:00
  • If we assume everybody votes, then an equivalent version of my answer is $\binom{m+n-1}{m-1}$. This is because in general $\binom{a}{b}=\binom{a}{a-b}$. If we don't assume everybody votes, we get answer $\binom{m+n}{n}$ or equivalently $\binom{m+n}{m}$. – André Nicolas Sep 01 '13 at 18:00
  • Yes, if not everybody votes, then m = m + 1 - we add a dummy candidate. And the problem is reduced to the previous problem. – user4035 Sep 01 '13 at 18:06
  • Don't forget the possibility that some people vote more than once :) – DJohnM Sep 01 '13 at 18:58
  • Early and often. I have seen it happen, though only at a nomination meeting. – André Nicolas Sep 01 '13 at 19:02
1

You're working way too hard here! This is a standard "stars and bars" problem.

You don't care who votes for specific candidates; just how many votes each receives. So, you are really trying to count up the number of ways of choosing non-negative integers $c_1,c_2,\ldots,c_m$ such that $c_1+\cdots+c_m=n$.

The number of ways to do this is $$ \binom{n+m-1}{m-1}. $$ Why? Any such arrangement can be written in the form $**|***|****||*$ or something similar, where there are $m-1$ lines and $n$ stars: the stars before the first bar correspond to votes for the first candidate; between the first and second for the second candidate; and so on. The number of such arrangements is the number of ways to choose, out of the $n+m-1$ positions total, which $m-1$ of them are bars.

Nick Peterson
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0

As no of voters is n, no of votes will also be n.

No of ways of selecting n votes among m distinct candidates, is $$ \binom{m+n-1}n \qquad\ $$

Sumedh
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