# combinatorics problem : How many different choice are there So that exactly 3 candidate get most vote.

I need to solve this problem about combination :

How many different choice are there in choice representative student group with five candidate and 40 students.(every student should be choose one candidate and one vote is just for one person )so that:

1)exactly 3 candidate get the maximum vote.

My attempt:

$x_1+x_2+x_3+x_4+x_5=40 \to x_1=x_2=x_3 \to \begin{cases} 3x_1+x_2+x_3=40 \\(x_1 \gt x_4 )and (x_1 \gt x_5) \end{cases}$

how to continue?

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What does "exactly 3 candidate get most vote" mean? – wj32 Dec 15 '12 at 10:32
@wj32 "exactly 3 candidate earn maximum vote" – Minsin Minsini Dec 15 '12 at 10:34
is it an open vote-- is it known or does it matter who voted for who ? – ashley Dec 15 '12 at 10:40
What exactly is the protocol? Can one vote fr ore than one person? You have to add some details. – Michael Greinecker Dec 15 '12 at 10:41
@MichaelGreinecker "every student should be choose one candidate" and "one vote just for one person " – Minsin Minsini Dec 15 '12 at 10:46

There are $\binom53$ ways to choose which $3$ candidates come equal first. Since $5\cdot8=40$, their score must be at least $9$, and it can obviously be at most $13$. Now make a little table:

$$\begin{array}{c|l} \text{Top vote}&\text{Remaining votes}\\ \hline 9&8,5;\quad7,6\\ 10&9,1;\quad8,2;\quad7,3;\quad6,4;\quad5,5\\ 11&7,0;\quad6,1;\quad5,2;\quad4,3\\ 12&4,0;\quad3,1;\quad2,2\\ 13&1,0 \end{array}$$

If the top vote is $9$, the remaining votes must have been split either $8,5$ or $7,6$. That’s $2$ possibilities, and each can happen in $2$ ways, depending on which of the remaining two voters gets the larger number of votes. Thus, there are $\binom53\cdot2\cdot2=40$ ways to get the vote totals $9,9,9,8,5$ and $9,9,9,7,6$, corresponding to the $40$ distinguishable permutations of those five numbers.

If the top vote is $11$, there are $4$ different ways to split the remaining $7$ votes into two parts, and each of those can be assigned to the other two candidates in either order, so there are $\binom53\cdot4\cdot2=80$ ways to get three vote totals of $11$ and two more adding up to $7$.

That’s two of the five cases; the other three can be done similarly, so I’ll leave them for you to try.

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Somehow you overlooked $8,5$ for case "9"? – Hagen von Eitzen Dec 15 '12 at 11:16
@Hagen: Thanks very much; fixed. (By the way, the English idiom is multiply by, not multiply with.) – Brian M. Scott Dec 15 '12 at 11:19
Thanks for nice way, Does answer is ${5 \choose 3}4+{ 5\choose 5}9+{5 \choose 3}8+{5 \choose 3}5+{5 \choose 3}2$ ? – Minsin Minsini Dec 15 '12 at 12:01
@Minsin: Yes, except that you have a typo in the second term: it should be $\binom539$. – Brian M. Scott Dec 15 '12 at 12:04

If invalid votes or abstentions don't occur, there are $5\choose 3$ ways to choose the winners $A,B,C$. If they obtain $k$ votes each, then $k\ge 9$ as otherwise one of the others ($D,E$) would have at least $8$ votes as well. And of course $k\le 13$. There are $40-3k$ votes to be split among the other two such that none of them gets $k$ or more votes.

• $k=9$: $13$ votes left, $D$ can obtain $5,\ldots,8$ votes, $E$ the rest
• $k=10$: $10$ votes left, $D$ can obtain $1,\ldots,9$ votes, $E$ the rest
• $11\le k\le 13$: $D$ can obtain $0,\ldots, 40-3k$ votes, $E$ the rest.

Sum up the possibilities and dont forget to multiply with $5\choose 3$.

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Thanks for your way. – Minsin Minsini Dec 15 '12 at 12:03