Number of possible results in election with one of candidates getting more then 50% votes I have been trying to solve a question in combinatorics which is as follows:
How many possible results are there for an election where 80 people voted for 4 different candidates where one of the candidates received more then 50% of the votes?
I have tried to figure out how to do this and can't seem to manage I started with calculating the entire possible results which came out to be $\frac{83!}{3!}$ and continue from there.
But I can't seem to find the way. I know I need to use the inclusive exclusive principle but I am not sure how to apply it.
Could anyone please help? 
Thanks a million.
 A: Since A must have at least 41 votes, go ahead and set the problem up likeA: 41, B: 0, C: 0, D: 0.
The problem now is how to distribute the remaining 39 votes among the 4 candidates where candidates might receive no votes. That is, we must arrange 39 unlabeled balls (votes) into 4 labeled boxes (candidates) where boxes may be empty.
In general, we can arrange $n$ unlabeled balls into $m$ labeled boxes with some possibly empty in
$$
\binom{n + m - 1}{n}
$$
ways. (See, for example, here if you'd like more on this result.)
Thus, for the problem at hand, we get
$$
\binom{42}{39}
$$
possible outcomes for the election.
EDIT: This solution assumes that candidate A is the winner. To account for the possibility of any candidate being the winner, multiply the previous result by 4:
$$
4 \binom{42}{39}.
$$
A: There are $n+1$ compositions of $n$ into two parts.  For compositions into three parts, let the first part be $i$ and sum over $i$, so there are $\sum_{i=0}^n (n-i)+1=n+1+\frac{n(n+1)}{2}=\frac{(n+1)(n+2)}{2}$ compositions of $n$ into three parts.  Now let the winner have $j$.  If the winner is the first part, there are $\sum_{j=41}^{80}\frac{(n-j+1)(n-j+2)}{2}$ compositions.  Now multiply by $4$ to move the winner to any location.
