You have two sets of four numbered wooden blocks. Each set contains blocks with the numbers 1,2,3 and 4.

You have to place all eight blocks in a tower so that there is precisely:

1.    one block between the two blocks numbered 1;
2.    two blocks between the two blocks numbered 2;
3.    three blocks between the two blocks numbered 3;
4.    four blocks between the two blocks numbered 4.

How many ways can you do this?

Please note, I actually have no sodding clue what the answer is, because I am a lazy sod. I just got a couple of mathsy questions from a friend and this one I looked at and thought "fuck it, put it on Tangler, they'll know how to work it out without brute-forcing it".

So... the first answer explained in words of one syllable or less gets a point 

i can do it one way

Brute forcing it might be quicker!

I will assume you cannot "cycle" from the top of the tower to the bottom, so e.g. one of the blocks numbered 1 being at the bottom of the tower, and the other being second from the top would NOT be a valid solutions.

Let us number the positions a,b,...,h, with a the bottom block in the tower and h the top block.

Then there are 3 possible positions for the blocks numbered 4, being {(a,f),(b,g),(c,h)}.

There are 4 possible positions for the blocks numbered 3, being {(a,e),(b,f),(c,g),(d,h)}.
By inspection, only 2 of these are compatible with each of the 3 possibilities for blocks numbered 4 above.
e.g. the (a,f) choice of positions for the blocks numbered 4, rules of the following choices for the blocks numbered 3: {(a,e),(b,f)}.
This gives 3+3 = 6 possible combinations for the blocks numbered 3&4.

Then there are 5 possible positions for the blocks numbered 2, being {(a,d),(b,e),(c,f),(d,g),(e,h)}.
By inspection, for each of the 3 possibilities for blocks numbered 4 above, for one of the compatible positions of the blocks numbered 3 above, only 2 of these possible positions for the blocks numbered 2 are compatible (e.g. the choice {{(a,f)},{(c,g)}} means only {(b,e),(e,h)} are allowed for the blocks numbered 2); and for the other compatible position of the blocks numbered 3 above, only 1 of these possible positions for the blocks number 2 is compatible (e.g. the choice {{(a,f)},{(d,h)}} means only {(b,e)} is allowed for the blocks numbered 2).
This gives 3*2 + 3*1 = 9 possible combinations for the blocks numbered 2&3&4.

Then there are 6 possible positions for the blocks numbered 1, being {(a,c),(b,d),(c,e),(d,f),(e,g),(f,h)}.
By inspection, for each of the 3 possibilities for blocks numbered 4 above, for the first 6 of the compatible positions of the blocks numbered 2&3 above, only 1 of these possible positions for the blocks numbered 1 are compatible; whereas for the last 3 of the compatible positions of the blocks numbers 2&3 above, NONE of these possible positions for the blocks numbered 1 are compatible.
This gives 3*2*1 + 3*1*0 = 6 possible combinations for the blocks numbered 1&2&3&4.

Actually I typed that on a train about an hour ago, but didn't have internet to submit it...

How did you read the question if you had no internet? 

Okay, I'm an addict. I checked for new questions before I headed off for the train, and did it on my computer on the train offline.

Just checking through your thing now... any last minute checking you want to do, now would be the time to do it 

hmm... following the same basic method as you, I'm coming up with a different final answer

Starting at step for blocks numbered 2.  It should be (1+2) + (1+1) + (2+1).

Then, for these, either one or zero possible valid choices for blocks numbered 1, as follows:

Total = (1*0+(1*1+1*0)) + (1*0+1*0) + ((1*1+1*0)+1*0) = 2.

Hmm, how did I screw that up first time around?

I'm still getting a different answer...

Taking it from the top, you've got 3 options to place the #4s and 2 options for the #3s for each of those.

Gives you the following options (listed from the top of the stack to the bottom)

1. 4x3xx43x
2. 4xx3x4x3
3. 34xx3x4x
4. x4x3xx43
5. 3x4x3xx4
6. x34xx3x4

For options 1 and 6 you can put the #2s in 2 places, for options 2, 3, 4, and 5 you can only put them in one place.

Giving you

1. 423x243x
2. 4x3x2432
3. 42x324x3
4. 34x2324x
5. x423x243
6. 3x423x24
7. 2342x3x4
8. x342x324

And only 2 of those options allow a valid placement for the #1s (and they're mirror images of each other)

Wait... dawning realisation that you can invert 2 blocks with the same number

Cock. Now I'm confused again.

Check my edited revision.

If you arrange the 4s as x4xxxx4x then it leads to only one placement of the 2s for each placement of 3s (neither of which have a place to put the 1s)

I have spreadsheet evidence! 

Check my edited revision.

I think that is finally it. And my battery is about to run out.

Hey, I see you also noticed the (b,g) choice for the blocks numbered 4 doesn't allow any solution. (Only one place to put the blocks numbered 2, nowhere to put the blocks numbered 1).

Also that for the way of arranging 3s that gives you 2 ways to put the 2s, only one of those placements gives you somewhere to put the 1s

Cuts it down to 2 options, if you ignore the possibility of swapping 2 of the same block over

in which case you get a doubling of possible answers for each type of block, and 32 solutions

I think 4 solutions. Whoops, no, I think it has 2.

1. 423x243x
2. 4x3x2432

Only number 2 can have the 1s separated by 1 block

I am assuming the two blocks numbered n are indistinguishable.

Remember mirror images ARE different solutions, because it is a tower, so has a top and a bottom.

I know.

But taking the 1st and 3rd way to arrange the 4s, each has 2 places you can put the 3s. One of those ways has no final solutions, the other has 2 places you can put the 2s. Only one of those ways gives you a valid answer.

One solution for the 1st way to put the 4s, one solution for the 3rd way to put the 4s.

If I was disallowing mirrors, I'd say there was only 1 answer...

I take it from your edit that I have you persuaded 

and

23421314

Yes, before I posted these, I edited my revised solution above.

Right... thanks for that.

I don't know what to do about points, since I kinda worked most of it out, but wouldn't have been able to do so without your first post.

Also SirM was actually kinda right...  (One solution, then mirrored)

Meh, half a point to ye for your help, deal?

Fair enough.

I should have written down the possibilities when I got to the third step and not tried to keep all of the disallowed combos in my head. i.e. brute force was pretty much the way forward.

HAHAHA!!

no wonder you like the brute force option...

Railguns are better.

10x better.

Well if you want to make a version of that picture with a rail gun in the middle, then be my guest 

I would have to counter with a picture of the Moon colliding with the Earth, but you go right ahead with your little rail gun.