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

 A: We need to determine the number $N$ of 3-tuples from this set of numbers that have the desired property that one number is the average of the other two (We also need the total number of 3-tuples from this set, but think isn't a challenging task).
Let's first think about what is needed for a number $n$ to be the average of two other numbers $a$, and $b$. Based on the definition of average, we have
$$n=\frac{a+b}{2}$$ 
A way to look at this equation is to say that $n$ is the midpoint of the interval $[a,b]$. Therefore, to find $N$ all we need to do is figure out how many intervals are possible for each number in our set, and then add them all up. 
First, we don't need to consider $1$ or $20$, because they do not have an interval around them for this set. For the other numbers, the number of intervals just depends on the closest distance that number is from either $1$ or $20$. More specifically, we need the smaller distance of the two.
Therefore, 
$$N=\sum_{n=2}^{19}\text{Min}(20-n,n-1)$$
But we can easily determine which one is the minimum for a given number. When $n\leq10$, $n-1$ is the minimum, and when $n\geq11$, $20-n$ is the minimum.
Therefore,
$$N=\sum_{n=2}^{10}(n-1)+\sum_{n=11}^{19}\text(20-n)$$
These sums should be easily manageable now. I will leave the rest of the work to you.
A: The number of 3 long arithmetic sequences is:
Let $a_d$ be a number of such sequences if difference is $d$ 
$d=1:\;\;\;$$(1,2,3);(2,3,4),...(18,19,20)$, so $a_1=18$. 
$d=2:\;\;\;$$(1,3,5);(2,4,6),...(16,18,20)$, so $a_2=16$.
$d=3:\;\;\;$$(1,4,7);(2,5,8),...(14,17,20)$, so $a_3=14$.
$\dots$
$d=9:\;\;\;$$(1,10,19);(2,11,20)$, so $a_9=2$.
So the number of all good sequences is $2+4+...+18 = 90$. Thus the answer is $$P= {90\over {20\choose 3}} = {90\over 20\cdot 19\cdot 3}= {3\over 38}$$
A: 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: 
$$\frac{90}{{20\choose 3}}=\frac{90\cdot 1\cdot 2\cdot 3}{20\cdot 19\cdot 18}=\frac{3}{38}.$$
