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There is this question. There are six different candidates for governor of a state. In how many different orders can the names if the candidates be printed on a ballot?

The answer is $6!=720$. But why?

If there are $6$ candidates and order matters, they are being placed on $1$ card. Wouldn't it be $p(6,1)=6$ ways?

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    $\begingroup$ p(6,1) is the number of ways to arrange 1 out of the 6 candidates. An election in which only one out of the six candidates ever gets their name printed on a card would be a joke. Note further that you should be able to write down much more than 6 ways to arrange 6 names, with no real effort right off the top of your head. This should be obvious! Hell, 6 is the number of ways you can arrange just three names. By definition, the number of ways to arrange $k$ out of $n$ possible things in some order is $p(k,n)$; here $k=6$ and $n=6$. $\endgroup$
    – anon
    Dec 14, 2014 at 23:48
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    $\begingroup$ Perhaps the confusion stems from not knowing what a ballot would look like. In an election, the ballot will have the names of all candidates in a list (in our case six), where the voter then gets to put a check mark, hole, or whatever method of identification is used in that region on the card next to the candidate they want to win. $\endgroup$
    – JMoravitz
    Dec 14, 2014 at 23:54
  • $\begingroup$ The "card" is just the place where we write the list of the candidates' names. In practice, there is typically only one list of items involved in a problem whose answer is $p(m,n),$ even if $n$ is a much larger number. $\endgroup$
    – David K
    Dec 15, 2014 at 0:11

3 Answers 3

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$P(6,1) = 6$ is the number of ways you can select a candidate appointed as a governor while

$6! = P(6,6) = 720$ is the number of ways you arrange the $6$ names on a given ballot.

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By the reasoning suggested in the question [which is incorrect, I must add, in order not to be misunderstood in the way in which I have already be misunderstood in a comment below.], if there were four candidates, named $A,B,C,D$, then there would be four orders in which they can be listed. Let's see if we can find all four: \begin{align} ABCD \\ ABDC \\ ACBD \\ ACDB \\ ADBC \\ ADCB \\[8pt] BACD \\ BADC \\ BCAD \\ BCDA \\ BDAC \\ BDCA \\[8pt] CABD \\ CADB \\ CBAD \\ CBDA \\ CDAB \\ CDBA \\[8pt] DABC \\ DACB \\ DBAC \\ DBCA \\ DCAB \\ DCBA \end{align} Now count them and see whether the answer is closer to $4$ or to $4!=24$.

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  • $\begingroup$ Sorry, I misunderstood. I apologize. $\endgroup$ Dec 15, 2014 at 10:43
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The given answer is correct.

You have $6$ choices for the first name.

For each of the $6$ choices, you have $5$ choices for the second name, and so on.

So, by the multiplication principle, we have that there are $6!$ ways to print the names.

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