Combinatorics. A box contains 20 balls numbered $1,2,3,...,20$. 
A box contains $20$ balls numbered $1,2,3,...,20$. If 3 balls are randomly taken from the box, without replacement, what is probability that one of them is the average of the other two?

Which one of these is the correct answer?


*

*7/45

*3/38

*6/155

*3/154


Refer to the table:
$$\begin{array}{c|c|c}
\text{Middle (average) number}&\text{Sets}&\text{Number of sets}\\
\hline
2&\{1,2,3\}&1\\
3&\{1,3,5\},\{2,3,4\}&2\\
4&\{1,4,7\},\{2,4,6\},\{3,4,5\}&3\\
\vdots&\vdots&\vdots\\
9&\{1,9,17\},\{2,9,16\},\cdots,\{8,9,10\}&8\\
10&\{1,10,19\},\{2,10,18\},\cdots,\{9,10,11\}&9\\
11&\{2,11,20\},\{3,11,19\},\cdots,\{10,11,12\}&9\\
12&\{4,12,20\},\{5,12,19\},\cdots,\{11,12,13\}&8\\
\vdots&\vdots&\vdots\\
18&\{16,18,20\},\{17,18,19\}&2\\
19&\{18,19,20\}&1\\
\hline
\text{Total}&&2\cdot \frac{(1+9)\cdot 9}{2}=90
\end{array}$$
Hence, the required probability is: 
$$\frac{90}{{20\choose 3}}=\frac{90\cdot 1\cdot 2\cdot 3}{20\cdot 19\cdot 18}=\frac{3}{38}.$$
