put 4 red, blue and green triangles in a row How many ways are there to put 4 red, 4 blue and 4 green triangles in a row so that no blue triangles would lie next to each other.
I got an answer, there are 126*56=7056 combinations. Could you check please, if the answer is right/wrong?
 A: The whole point is that after choosing the order of red and green triangles (and you correctly found out that there are $\frac{1.2.3.4.5.6.7.8}{1.2.3.4.1.2.3.4}=\frac{8!}{4!4!}=\binom 84=70$ possibilities), then there are 9 positions, where you can put the blue triangles.

To choose 4 positions out of 9, you have $\binom94=126$ possibilities.
So the result is $70\cdot126=8820$.
NOTE: This was already explained in Brian's answer. I've decided to add a picture, since it might be useful for some users.
A: HINT: Do the problem in two steps.


*

*First determine in how many ways the red and green triangles can be arranged in a row. 

*Each of those ways is a row of eight triangles; counting the spaces at the beginning and end of the row, there are nine slots into which you can insert a blue triangle. Since you’re not allowed to have two blue triangles in a row, you can put at most one blue triangle into a slot. How many ways are there to choose four of the nine slots to be filled with blue triangles?
Now what must you do with the answers to (1) and (2) to answer your problem?
A: So, first things first We need to find how many ways are there to put 4 red and 4 green triangles in a row. It'll be something like this RGRGRG and so on. For example we have different triangles. That is why firs triangle is one of 8, second is one of 7... And so on.
The number of combinations is $8*7*6...*3*2*1 = 40320.$
Then we need to consider that some combinations are just place-switching, for example we put R1 instead of R2, but really nothing changes, because they all are red. And the same thing for greens.
That is why we need to divide that thing:
$(8*7*6*...*3*2*1)/(4*3*2*1)$ - Excluding reds
$(8*7*6*...*3*2*1)/(4*3*2*1)/(4*3*2*1)$ excluding both - red and green, which just switch places.
Next I should do something with blue ones.
Well, I can include them into considerations, and I think I can count the number of combinations (it will be something like this? $12*11*10...3*2*1$)
But I also need to exclude the number where two blues stand next to each other. Will it be a half of all combinations?
Sorry, but I'm stuck again. Could you help me?
A: total there are $\frac{12!}{(4!)^3}$ combination no.of combination and  the blue triangle will lie next to each other =$\frac{11!}{3!4!4!}$ in conclusion there will be 23100
