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I am having trouble understanding permutations. I have a better time understanding them visually.

Say I have total of $5$ fruit and two baskets. Basket A can only hold $3$ fruit and the Basket B can only hold $2$ fruit. How do I solve this permutation?

Is the idea that with Basket A, there are 5 options and with Basket B, there are 2 remaining options, so is the way to get the permutations $5\cdot2=10$?

After looking over other problems, I think the answer is $5 \choose 3$, but I have just memorized this and I am really trying to understand why it is the answer.

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  • $\begingroup$ This is actually a combination problem since the order of selection does not matter. $\endgroup$ – N. F. Taussig Feb 28 at 10:45
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Consider filling Basket A first. For this, you have five fruit and must choose three of them. There are 5 choices for the first fruit, and once this is chosen there are 4 for the next fruit and finally 3 choices for the last fruit and so $5\cdot 4\cdot 3$ choices. Since the order of the fruit going in to the basket is irrelevant one must divide through the number of ways of ordering the 3 fruit to avoid overcounting. There are $3!$ ways to order these three fruits (namely the number of permutations of three elements) and so there are \begin{align*} \frac{5\cdot 4\cdot 3}{3!} = \frac{5!}{3!2!}=\begin{pmatrix} 5 \\ 3\end{pmatrix} \end{align*} choices. For Basket B, there are only two remaining fruit and both must go into the basket and so there are no extra choices and so this is the final answer.

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  • $\begingroup$ If we cared about the order of the fruit going in the baskets, would we then do something with Basket B or would there still be no extra choices? $\endgroup$ – Sam Feb 27 at 21:27
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    $\begingroup$ Yes then we would need to multiply by 2 to account for the two possible permutations of putting these two fruits into the basket (and of course, not divide through by the $3!$ when putting the fruit into basket A). $\endgroup$ – user504775 Feb 27 at 21:30
  • $\begingroup$ In the case of order mattering, why is there no need to do permutations with the order the fruit go into Basket B? Is the idea that there are 3 types of permutations :1- Between Basket A and B (and we only care about the order they go into Basket A), 2- Within Basket A, 3-Within Basket B) $\endgroup$ – Sam Feb 27 at 21:33
  • $\begingroup$ We do care about their order: after allocating the fruit to basket A, there are two choices left for the first fruit to go into basket B and 1 choice for the last fruit, if that's what you meant? Alternatively, you could start with basket B: there are 5 choices for the first fruit to go in and 4 for the second, then onto basket A, there are a further 3 choices for the first fruit, 2 for the second and 1 for the final fruit so you end up with the same answer $5!$ when order matters. $\endgroup$ – user504775 Feb 27 at 21:39
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How many possibilities do you have for the basket $A$ ? You have to choose $3$ fruits among a total of $5$ fruits. By definition, the number of possibilities is ${5 \choose 3}$.

Now, once you have choosen your $A-$fruits, then you have no more choice for the basket $B$, you have to put the two others fruits in it. So you are done.

Finally the total number of possibilities is $${5 \choose 3}$$

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