Mixed order and unordered counting If you care about ordered or unordered counts you can use nCr or nPr. But If you want a few in a certain order and a few unordered is there a general approach?
For example imagine you have 6 red balls and 5 blue balls and you select 3 balls.
How large is the set of balls where the first is blue and the remaining two are red.
I determined it to be 25. 
If order didn't matter you could have (6 red choose 2 red) * (5 blue choose 1)=75
Then by symmetry, a third must start with blue. So 75/3=25
RRB
BRR
RBR
I'm wondering if there is a more general way to think about this problem.
 A: Unordered. Let's look at this for the case where order doesn't matter.
Altogether there are ${11 \choose 3} = 165$ ways in which to choose 3 balls from among 11 without regard to order.
As you say there are five ways to pick the blue ball
and ${6 \choose 2} = 15$ ways to pick the two red balls.
So, if order doesn't matter, the probability of getting one blue ball and two red balls is $75/165 = 0.4545.$ 
Ordered. First, we can get the answer from above. As you say, if we randomly arrange the unordered choices, then 1/3 of them will have the blue ball first. So that's probability $0.4545/3 = 0.1515.$ 
Second, we can work it directly as an ordered problem: Pick the blue ball with probability $5/11.$ Then pick two red balls (undistinguishable order) with probability $\frac{6}{10}\cdot\frac 5 9 = \frac 13.$ So the probability of BRR is $5/75 = 0.1515.$
Simulation for unordered samples.  To make counting easy, let blue balls be represented by 0s an red balls by 1s. The following R code does a simulation of a million experiments choosing 3 balls, and finds the
proportion of outcomes with two red balls. With a million iterations we can expect two or three place accuracy.
set.seed(720)    # For reproducibility
urn = c(0,0,0,0,0,1,1,1,1,1,1)
nr.red = replicate(10^6, sum(sample(urn,3)))
mean(n.,red==2)
[1] 0.454439

The number of red balls chosen at random from the urn without replacement has a hypergeometric distribution.
The histogram below shows simulation results, the red dots show exact hypergeometric probabilities.
hdr="Numbers of Red Balls Chosen"
hist(nr.red, prob=T, br=(-1:3)+.5, col="skyblue2", main=hdr)
 x=0:3; pdf=dhyper(x, 6, 5, 3)
 points(x,pdf,pch=19,col="red")


