14. A class consists of pupils who each wear either a red tie or a green tie. There are two more pupils wearing green ties than wearing red ties. Two pupils are chosen at random. The probability of selecting two pupils of different tie-colour is exactly treble the probability of selecting two pupils who both wear red ties. Given that $\mathrm{R}$ is the number of pupils wearing a red tie, and that $R>0$, determine $R^{3}+R^{2}+R$.
SOLUTION
399

The number of pupils wearing a green tie is $R+2$, and the total number of pupils in the class is $R+R+2=2 R+2$. Let the probability of selecting two red-tie wearing pupils be $q$.
Then $\frac{R}{2 R+2} \times \frac{R-1}{2 R+1}=q$.
Also, $\frac{R}{2 R+2} \times \frac{R+2}{2 R+1}+\frac{R+2}{2 R+2} \times \frac{R}{2 R+1}=3 q$.
Eliminating $q$ we have $3\left(\frac{R}{2 R+2} \times \frac{R-1}{2 R+1}\right)=\frac{R}{2 R+2} \times \frac{R+2}{2 R+1}+\frac{R+2}{2 R+2} \times \frac{R}{2 R+1}$.
Therefore, $3 R(R-1)=2 R(R+2)$ and hence $3 R^{2}-3 R=2 R^{2}+4 R$.
This quadratic equation simplifies to $R^{2}-7 R=0$ and has solutions $R=0$ and $R=7$.
We are given that $R>0$, so $R=7$ and $R^{3}+R^{2}+R=343+49+7=399$.