Permutations: Dividing 5 pieces of fruit between 2 baskets

Question

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.

Answer 1

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.

Answer 2

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 $${5 \choose 3}$$