If $\;$ ${^{56}}{}{P}_{\left(r + 6\right)} : {^{54}}{}{P}_{\left(r + 3\right)} = 30800 : 1$, find $r$.
Given: $\;$ ${^{56}}{}{P}_{\left(r + 6\right)} : {^{54}}{}{P}_{\left(r + 3\right)} = 30800 : 1$
i.e. $\;$ $\dfrac{56!}{\left(56 - r - 6\right)!} : \dfrac{54!}{\left(54 - r - 3\right)!} = 30800 : 1$
i.e. $\;$ $\dfrac{56!}{\left(50 - r\right)!} \times \dfrac{\left(51 - r\right)!}{54!} = \dfrac{30800}{1}$
i.e. $\;$ $\dfrac{56 \times 55 \times 54!}{\left(50 - r\right)!} \times \dfrac{\left(51 - r\right) \left(50 - r\right)!}{54!} = 30800$
i.e. $\;$ $56 \times 55 \times \left(51 - r\right) = 30800$
i.e. $\;$ $51 - r = \dfrac{30800}{56 \times 55} = 10$
i.e. $\;$ $r = 51 - 10 = 41$