If ${^{8}}{C}_{r} - {^{7}}{C}_{3} = {^{7}}{C}_{2}$, find $r$
Given: $\;$ ${^{8}}{C}_{r} - {^{7}}{C}_{3} = {^{7}}{C}_{2}$
i.e. $\;$ $\dfrac{8!}{\left(8 - r\right)! \times r!} - \dfrac{7!}{\left(7 - 3\right)! \times 3!} = \dfrac{7!}{\left(7 - 2\right)! \times 2!}$
i.e. $\;$ $\dfrac{8!}{\left(8 - r\right)! \times r!} = \dfrac{7!}{4! \times 3!} + \dfrac{7!}{5! \times 2!}$
i.e. $\;$ $\dfrac{8!}{\left(8 - r\right)! \times r!} = \dfrac{7!}{4! \times 2!} \left(\dfrac{1}{3} + \dfrac{1}{5}\right)$
i.e. $\;$ $\dfrac{8!}{\left(8 - r\right)! \times r!} = \dfrac{8 \times 7!}{4! \times 2! \times 3 \times 5}$
i.e. $\;$ $\dfrac{8!}{\left(8 - r\right)! \times r!} = \dfrac{8!}{5! \times 3!}$
i.e. $\;$ $\left(8 - r\right)! \times r! = 5! \times 3!$
$\implies$ $r! = 3!$ $\implies$ $r = 3$