Hexagon Numbering Problem 
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?

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.
