# Permutations: Dividing 5 pieces of fruit between 2 baskets

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.

• This is actually a combination problem since the order of selection does not matter. – N. F. Taussig Feb 28 at 10:45

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.
• 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). – user504775 Feb 27 at 21:30
• 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. – user504775 Feb 27 at 21:39
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 $${5 \choose 3}$$