What is the mean of these twelve mixed numbers?

Question

How to solve the problem without trying one by one? I want to improve my knowledge in this area. So what should I read?

There are twelve different mixed numbers that can be created by substituting three of the numbers 1, 2, 3 and 5 for a, b and c in the expression $a{b\over c}$, where $b<c$. What is the mean of these twelve mixed numbers? Express your answer as a mixed number.

Answer 1

These numbers are: $1 + \frac{2}{3}, 1 +\frac{2}{5}, 1 + \frac{3}{5}, 2+\frac{1}{3}, 2+\frac{1}{5},2 + \frac{3}{5}, 3+\frac{1}{2}, 3 + \frac{1}{5}, 3+\frac{2}{5}, 5+\frac{1}{2},5+\frac{1}{3}, 5+\frac{2}{3}$

Let their sum be $S$ and their mean $\mu$, then $\mu = \frac{S}{12}$

You could and probably should add up the whole numbers and the fractions serparately, but I chose to do it this way.

$S = \frac{5}{3} + \frac{7}{5} + \frac{8}{5} + \frac{7}{3} + \frac{11}{5} + \frac{13}{5}+\frac{7}{2} + \frac{16}{5} + \frac{17}{5} + \frac{11}{2} + \frac{16}{3} + \frac{17}{3}$

$S = \frac{50 + 42 + 48 + 70 + 66 + 78 + 105 + 96+102 + 165+160 + 170}{30} $

$S = \frac{192}{5}$

Then $\mu = \frac{192}{5} \cdot \frac{1}{12} = \frac{16}{5} = 3 + \frac{1}{5}$

Answer 2

When focusing on $a$, for any number in your $\{1,2,3,5\}$, you can create three pairs from the other remaining three numbers. Hence, when calculating your average, you will calculate average of $3\cdot1$ plus $3\cdot2$ plus $3\cdot3$ plus $3\cdot5$. In other words, average of three times sum of your numbers.

When focusing on $\frac{b}{c}$, since $b<c$, there are six pairs one can create out of your $\{1,2,3,5\}$ and you want to calculate average of twice of each fraction. That is, $\frac{1}{12}\cdot2\cdot(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{2}{3}+\frac{2}{5}+\frac{3}{5})$. The long sum is $\frac{27}{10}$.

You should read nothing, this is just adding numbers.