Unique Combinations I hope someone can help me with some combinations (and perhaps permutations).
This is still the hardest area of math for me, but I'm still trying.
This is a two part question.
(1)
I have a bag of apples (A) and a bag of bananas (B), I would like to find out in how many different ways I can pick 5 different fruits.
Order is important, so AABBB is different from ABBBA.
(2)
Let's now say I have to pick 3 apples and 2 bananas, how many unique ways can I do this?
I know this will be a subset of the result sets in question 1, but I don't know how to find the answer.
I've been searching for an answer to this, but I keep getting answers to when the amount of picks are fewer than the selections, which is opposite to this, where I need to pick 5 from only 2 choices.
Thanks in advance.
 A: 1: For each slot 1 to 5 we have 2 options, and they are all independent of another. The total number of configurations is therefore $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^5 = 32$.
2: Here you want to use combinations. Basically, imagine you're drawing 3 positions (without replacement, order doesn't matter) for the apples from a bag with labels $\{1, 2, 3, 4, 5\}$. The number of different configurations there equals the number of different ways to draw 3 apples and 2 bananas, and equals ${5 \choose 3} = 10$.
A: An effort to give you some understanding about the second case. 
Let the apples have the numbers $1,2$ and let the bananas have the numbers $3,4,5$. These $5$ numbers can be arranged in $5!$ ways. However, if we compare the possibilities $12345$ and $21456$ then both stand for $AABBB$. You could say that combination $AABBB$ will be counted this way more than once. How many times is it counted then? Note that $12$ and $21$ both result in $AA$ (there are $2!=2$ possibilities) and that $345$, $354$, $435$, $453$, $534$ and $543$ all result in $BBB$ (there are $3!=6$ possibilities). That means that $AABBB$ is counted $2!3!=12$ times. This is also the case for any other combination. So to find the number of "essentially different" arrangements you should divide $5!$ by $2!3!$. 
More generally if there are $n_i$ "i-fruits" for $i=1,\dots,k$ then the correct answer will be: $$\frac{(n_1+\cdots+n_k)!}{n_1!\times\cdots\times n_k!}$$
A: 1) Clearly there's symmetry, so the number of ways to chose 1 banana and 4 apples and 4 bananas and 1 apples is the same. Now use binomial coefficients. The answeer should be $2^5$. 
2) Use binomial coefficients.
A: Suppose for a second that your have A(1), A(2), A(3), B(1), and B(2), where all the apples and banana's are distinguishable.
It's easy to see that to order all of these in any way, you have 5!(5 for the first place, 4 for the second, etc). We then divide by 3! and 2! because they really aren't distinguishable, so we're overcounting a number of cases. But this gets 10 ways to order 3 apples and 2 bananas. 
I remember I was taught this method, so how come other people get 32? Is there something wrong with my logic?
