14. Three different numbers are chosen at random from the list $1,3,5,7,9,11,13,15,17$, 19. The probability that one of them is the mean of the other two is $p$. What is the value of $\frac{120}{p}$ ?
SolUtion
720

There are $\frac{10 \times 9 \times 8}{3 \times 2 \times 1}=120$ equally likely choices of three numbers.
It is not possible for 1 or 19 to be the mean.
For 3 to be the mean there is only one possible selection, namely $1,3,5$.
Similarly 17 can only be the mean if $15,17,19$ is selected.
5 is the mean of both 1,9 and 3,7 ; while 15 is the mean of both 11,19 and 13,17 .
7 is the mean of 1,$13 ; 3,11$; and 5,9 ; while 13 is the mean of 7,$19 ; 9,17$; and 11,15 .
Finally, 9 is the mean of 1,$17 ; 3,15 ; 5,13$; and 7,11 ; while 11 is the mean of 3,$19 ; 5,17 ; 7,15$; and 9,13 .
There are 20 such selections in total, so $p=\frac{20}{120}=\frac{1}{6}$. Therefore, $\frac{120}{p}=120 \times 6=720$.