Find $n$ if ${^{n}}{C}_{\left(n - 4\right)} = 70$
Given: $\;$ ${^{n}}{C}_{\left(n - 4\right)} = 70$
i.e. $\;$ $\dfrac{n!}{4! \times \left(n - 4\right)!} = 70$
i.e. $\;$ $\dfrac{n \times \left(n - 1\right) \times \left(n - 2\right) \times \left(n - 3\right) \times \left(n - 4\right)!}{\left(n - 4\right)!} = 4! \times 70$
i.e. $\;$ $n \times \left(n - 1\right) \times \left(n - 2\right) \times \left(n -3\right) = 4 \times 3 \times 2 \times 70$
i.e. $\;$ $n \times \left(n - 1\right) \times \left(n - 2\right) \times \left(n -3\right) = 4 \times 3 \times 2 \times 5 \times 2 \times 7$
i.e. $\;$ $n \times \left(n - 1\right) \times \left(n - 2\right) \times \left(n -3\right) = 8 \times 7 \times 6 \times 5$
$\implies$ $n = 8$