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So here is Question :-

A triangular grid of side $n$ is formed from $n^2$ equilateral triangles with sides of length $1$.Determine the number of parallellograms.

First of all , by reading the question I can understand that the answer should be evaluated by some kind of counting or combinatorial shortcut . But I really don't know what formula I can use to find the no. of parallelograms of a triangular grid of side $n$. Can anyone help with some explanation ?

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3 Answers

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Separate the parallelograms into the 3 directions that they point in. Focus on a single direction.

Hint: Extend the setup by 1 more row.

For a given parallelogram, extend the 4 edges till they hit the extended row. This determines 4 unique points.
This is illustrated by the red parallelogram, whose edges are extended.

Conversely, given these 4 points, we can reconstruct the parallelogram by following the edges.
E.g. For the 4 yellow dots, which parallelogram do they determine?

Hence, there are $ 3 \times { n + 2 \choose 4 } $ parallelograms.

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So we have $1+2+...+n+(n+1)= {n+2\choose 2}$ vertices determined by this grid.

Any pair, which is not on the same line determined by this grid, determine opposite vertices of some paralellogram and any parallelogram is determined by exactly one such pair.

The number of bad pairs is $$3\cdot \Big({1\choose 2} + {2\choose 2}+...+{n+1\choose 2}\Big) = {n(n+1)(n+2)\over 2}$$

So the number of good pairs is = the number of paralleograms $$ {{n+2\choose 2}\choose 2} - {n(n+1)(n+2)\over 2} =\boxed{{(n-1)n(n+1)(n+2)\over 8}} $$

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The parallelograms can be built in three orientations choosing (see picture) segments lying on blue-red, red-black or black-blue pairs of lines. From symmetry it is enough to count the number of parallelograms for one coloring (say blue-red).

Start the counting with the upmost red line of length 1. There is only one way to choose a pair ob blue lines and $n-1$ ways to choose the other red line. All together we have $(n-1)$ ways to construct a parallelogram. Taking the next red line (with the length 2) we have $\binom 32$ ways to choose the blue lines and $n-2$ ways to choose the other (lower-lying) red line.

Continuing in this way we find that the overall number of blue-red parallelograms is:$$ \sum_{k=1}^{n-1} \binom{k+1}2(n-k)=\frac{(n-1)n(n+1)(n+2)}{24}. $$

The total number of the parallelograms is triple of this:$$\frac{(n-1)n(n+1)(n+2)}{8}. $$

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