Not so constant sum?

Let $a, b, c, d$ , and $e$ be the five numbers. Hence $\begin{aligned} (a+b+c)+(a+b+d)+(a+b+e)+(a+c+d)+\dots + (c+d+e) & = 15+15+15+\dots+15 \\ & = \binom{5}{3} \times 15 \\ & = 150 \\ 6a+6b+6c+6d+6e & = 150 &\text{Note that} \binom{4}{2}=6 \\ \ \\ \Longrightarrow a+b+c+d+e & = \dfrac{150}{6} = \boxed{25} \end{aligned}$

Here is an example: $a=b=c=d=e=5$ .