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A "normal" domino piece has two faces or numbers on each side, but this problem requires using pieces with 3 faces.

Furthermore, a (x,y,z) block is considered to be identical to a (z,y,x) block owing to their symmetry.

I know that 28 different domino blocks can be made from 2-faced blocks when using "n" different numbers.

I tried setting up multiple blocks and extending the value on each face to "n"; however, i can't find a pattern that can help me solve the problem.

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Sam202
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  • hint pascal's triangle. –  Sep 03 '17 at 23:57
  • The reason there are $28$ domino blocks is that each domino pairs two not necessarily distinct numbers from the set ${0, 1, 2, 3, 4, 5, 6}$, so there are $\binom{7}{1} + \binom{7}{2}$ dominoes. – N. F. Taussig Sep 04 '17 at 00:01
  • So if i'm dealing with 3-faced domino blocks, does that mean the total number of different blocks is 28 + 7 nCr 3 = 63? – Sam202 Sep 04 '17 at 00:06

1 Answers1

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There are $${\small{\binom{n}{1}+\binom{n}{2}}}$$ ways to choose the unordered outer pair (same as the number of $2$-faced dominoes).

For each such outer pair, there are $n$ possible middle values.

So the count you want is just the product $$\left(\small{\binom{n}{1}+\binom{n}{2}}\right)n$$

quasi
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