Arrangement Question I'm trying to represent the number of ways to arrange n blue and 4 red balls whose colors are indistinguishable
I've come up with, $${{n+4} \choose 4}$$ 
but I'm not sure how it works... How come I choose 4 but not n, and how come i am not permuting anything, since I'm dealing with arrangements?
 A: First, note that $\dbinom{n+4}{4}=\dbinom{n+4}{n}$, so "choosing $n$" would give the same answer. 
In general, $\dbinom{m}{k}=\dbinom{m}{m-k}$. Rows of Pascal's Triangle read the same forwards as backwards. If we wish, we can verify this by using the formula for binomial coefficients in terms of factorials. 
But there is a better way. There are exactly as many ways of picking $k$ "winners" from $m$ people as there are of picking $m-k$ "losers."
Second, yes, we are counting permutations, but binomial coefficients, which are useful in counting certain kinds of combinations, have many other uses.
To justify the answer $\dbinom{n+4}{4}$, imagine that we have $n+4$ small boxes in a row. There are $\dbinom{n+4}{4}$ ways to choose the boxes that the red marbles will occupy.  Probably that you had an argument like this in mind.
To put it another way, suppose we want to make an $(n+4)$-letter "word," using $4$ R's  and $n$ B's. There are $\dbinom{n+4}{4}$ ways to choose the position of the R's. Once we have chosen where the R's will go, the whole word is determined. 
A: The symmetry of the binomial coefficients tells you that ${{n+4} \choose 4}={{n+4} \choose n}$ so you can choose $n$ if you like.  Your argument is one way to see this symmetry.  In fact ${{n+4} \choose 4}=\frac {(n+4)!}{n!4!}$ which makes the symmetry evident.
