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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?

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Following Brian's hint I'm getting $\frac{8!}{4!4!}=70$ and $\binom94=126$. You wrote (in your last edit) 126 and 56, could you explain, how did you get those numbers? I'm guessing 126 is from the binomial coefficient, since we have the same result. – Martin Sleziak Jan 18 '12 at 11:22
Unfortunately, I'm not very good at theory, so I'll try to explain exactly like I understand this theme. – user1131662 Jan 18 '12 at 12:59
As soon as we both get 126, I'll explain just 56. There are 8 triangles. We pick one spot for a triangle. We have now seven. We pick a spot again. Now we have 6. And so on. So here we get $8*7*...2*1$ But we should exclude some triangles, because switching their places will not affect our triangle system. we exclude $4*3*2*1$ for reds and $4*3*2*1$ for greens. That's how I got the second number, is it right? – user1131662 Jan 18 '12 at 13:08
That's correct, except when I calculate it, I get 70 and not 56:… – Martin Sleziak Jan 18 '12 at 13:45
Oh, thank ypu very much, I'm really sorry for this stupid mistake. – user1131662 Jan 18 '12 at 15:20
up vote 2 down vote accepted

The whole point is that after choosing the order of red and green triangles (and you correctly found out that there are $\frac{}{}=\frac{8!}{4!4!}=\binom 84=70$ 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.

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Thank you very much, I think picture might be very helpful, even though I managed to understand it. – user1131662 Jan 18 '12 at 15:22

HINT: Do the problem in two steps.

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

  2. 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?

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Okay, starting with the first step - here I should write the number of all possible combinations. Am I right? – user1131662 Jan 13 '12 at 17:11
Oh, sorry, I understood the problem. – user1131662 Jan 13 '12 at 17:12
@user1131662: If you mean the number of distinguishable sequences of four red and four green triangles, yes; otherwise, no. Think of it this way. You’re going to lay out a row of eight triangles, four of them red. Once you’ve decided which four positions get red triangles, the green ones have to go in the remaining four slots. In how many ways can you pick four of the eight slots for the red triangles? – Brian M. Scott Jan 13 '12 at 17:15
Yes, that is what I am talking about. Then I pick four positions for red ones. And so I decide where to put these red triangles. Later I put the green ones among the red ones. The number of different combinations is my first goal. Am I right? – user1131662 Jan 13 '12 at 18:13
@user1131662: Yes, so your answer at this stage should be a single binomial coefficient. – Brian M. Scott Jan 13 '12 at 18:17

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?

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Referring to Brian's answer, you've done step 1 correctly - that is, you've successfully counted the ways to arrange just the red and green triangles. For step 2, you need to place four blue triangles into the nine spaces between red/green triangles such that you put at most one in each space. How many ways are there to choose four spaces for the blue triangles out of nine spaces total? – Lopsy Jan 16 '12 at 17:43
Well, if I did my self-training correctly, and I clearly understand your comment, total № of combinations will be $(9*8*7...*3*2*1)$. But I need to exclude a number of $(4*3*2*1)$ combination. Now I tried to do the thing I did previously with picking 4 slots for red triangles out of 8. – user1131662 Jan 16 '12 at 17:55
Is it right? I'm not sure unfortunately.. – user1131662 Jan 16 '12 at 18:52

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

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$23100$ is quite a bit larger than the correct answer. – Brian M. Scott Jan 13 '12 at 17:18
what is the answer you have manipulate? – Mathematics Jan 13 '12 at 17:24
I’d prefer not to post it until the OP has had more of a chance to think about my answer, but it’s less than $10,000$. You should be able to work it out from my answer above. – Brian M. Scott Jan 13 '12 at 17:31
did you find any mistake in the step?I thought it is just multinomial calculation – Mathematics Jan 13 '12 at 17:44
Your analysis is incorrect: there are more than $\frac{11!}{3!4!4!}$ arrangements that have two blue triangles next to each other. I’m not even sure why you picked that number. – Brian M. Scott Jan 13 '12 at 17:48

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