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If there are three people and three apples, what is the max amount of arrangements, assuming everyone gets an apple to themselves?

I've tried counting the different arrangements and got six but I feel like I could be missing a few. Is there any way to do this without counting?

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    You need to be a bit clearer as to what is distinct and what is indistinct and how many of what there is. If there are three people (*distinct*) and three *identical* red-delicious apples, the answer to how many ways we can distribute these apples so that everyone gets at least one apple is $1$. If the three apples are distinct as well and only one of each apple exists, (*e.g. you have one red-delicious, one granny smith, and one braeburn to distribute*) then the answer is $3!=6$. If you have multiple of each variety of apple available, it changes again... – 2017-02-28

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First person has three choices, the second has two and the third has only one choice, so $1\times 2\times 3$.

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    Thanks for the answer. Don't know how I forgot this – 2017-02-28
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There are at least four scenarios here, each with different assumptions, so each has a different answer.

  1. If there are 3 apples of the same type, there is only 1 way to distribute 3 apples among 3 people so that each person has exactly 1 apple.

  2. If there are 3 different types of apples (or 3 different fruits altogether), there are $3!=6$ different ways to distribute the apples/fruit so that each person has exactly 1 apple.

  3. If there are 3 bins of apples, each of a different type, and each bin has at least 3 apples in it, and each person gets only 1 apple, there are $3^3=27$ ways to distribute the apples.

  4. If it does not matter how many apples each person gets, from 0 to 3, and there are 3 identical apples to choose from, there are at least 10 ways to distribute the apples among 3 people.

To do this without counting requires formulas. The formulas should be as follows (numbered according to the scenarios above, where $a$ is the number of apples and $p$ is the number of people). Note that since in your original scenario $a=p$, it is difficult to discern between $a$ and $p$ in the formulas.

  1. $a \choose p$

  2. $p!$

  3. $a^p$

  4. I don't know of a formula to use here. I just counted.

Hope this helps.

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    Thanks for the in-depth answer. It helps a lot – 2017-02-28