Hexagon Numbering Problem

Question

So in the above hexagon figure, I have to arrange 1 to 7, inclusive, into the circles such that the three dark red triangles have the same sum. How many distinct arrangements can there be?

Answer 1

First find which numbers will be common for all shaded triangles (center circle) so the remaining numbers can be ordered in three pairs with the same sum.

If you put $\boxed{1 \Rightarrow 2+7=3+6=4+5}$

If you put $2$ there is no possible arrangament

If you put $3$ there is no possible arrangament

If you put $\boxed{4 \Rightarrow 3+5=2+6=1+7}$

If you put $5$ there is no possible arrangament

If you put $6$ there is no possible arrangament

If you put $\boxed{7 \Rightarrow 1+6=2+5=3+4}$

So in the center you can put $3$ numbers, then you select one number for the first saded triangle $\binom{6}{1}$, then by the second $\binom{4}{1}$ and last by the third $\binom{2}{1}$.

Therefore our result is $$3\times\binom{6}{1}\binom{4}{1}\binom{2}{1}=72$$

Note that I only select one number by triangle because the other one is definite by the first.