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How to align two boxes in a line?

You have five boxes in a line, each can be occupied by a 1, 2, or 3. Two adjacent boxes can also have 0s; an individual box, three in a row, or all the boxes can't be 0. What is the total number of unique combinations you can make? Also, what if you didn't have the 0 to worry about?

  • You can repeat 1, 2, and 3. Also this isn't homework, (it's summer break) this just popped into my head and I got curious about it

  • Answer:

    Note: This is answering an earlier version of the question: one where only sequences that contain three consecutive 0s were forbidden. The question was edited after I wrote the answer. 984. This is easy to count using the principle of inclusion and exclusion: There are 5 boxes, and 4 possibilities for each of them, for a total of [math]4^5=1024[/math] different sequences. From those, we have to subtract the forbidden ones. For each of the patterns "000--", "-000-", and "--000" we have [math]4^2=16[/math] forbidden sequences. We need to subtract those from the total. Of course, subtracting 3*16 is not completely correct. Why? Because we subtracted each of the four sequences that matches "0000-" twice, and the same is true for sequences that match "-0000". To make up for that, we add 2*4. As a final step, verify that the sequence "00000" is currently (correctly) counted zero times. Thus, the answer is: 1024 (all possible sequences) - 3*16 (subtract forbidden sequences) + 2*4 (fix the count for sequences that were subtracted twice) = 984.

Michal Forišek at Quora Visit the source

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It's not clear whether or not you allow 4 in a row boxes filled with 0s, so I'll answer the question both ways, but let's start with the no 0 case... case 1: no 0s allowed _ _ _ _ _ filled in with {1,2,3} => 3^5 = 243 possibilities case 2: allow 0's only in adjacent pairs, but no more than 2 in a row. 243 possibilities from case 1 plus... 0 0 _ _ _ or _ 0 0 _ _ or _ _ 0 0 _ or _ _ _ 0 0 filled in with {1,2,3} => 4*3^3 = 108 possibilities... plus 0 0 _ 0 0 filled in with {1,2,3} => 3 possibilities = 243 + 108 + 3 = 354 possibilities case 3: allow 0's only in adjacent pairs or quadruplets 354 possibilities from case 2 plus... 0 0 0 0 _ or _ 0 0 0 0 filled in with {1,2,3} => 6 possibilities = 354 + 6 = 360 possibilities

Andrew Weimholt

Assuming you can't repeat the numbers (1,2,3) then this is a permutations problem with a little twist. I won't do your homework, but here are some hints. There are two cases, one where there is one group of adjacent 0s and the other where there is two. You can think of the two zero boxes as one box, so for the first case: 1 2 3 00 1 2 00 3 1 3 2 00 1 3 00 2 ... On top of those permutations, you also have the other case where you have two groups of adjacent zeros. Again think of those adjacent zeros as one box. 1 00 00 2 00 00 3 00 00 00 1 00 ... Hope this helps.

Victor Aguilar

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