Solve the equation: $\;$ $C^{n+1}_{m+1} : C^{n+1}_{m} : C^{n+1}_{m-1} = 5 : 5 : 3$
$C^{n+1}_{m+1} : C^{n+1}_{m} : C^{n+1}_{m-1} = 5 : 5 : 3$
i.e. $\;$ $C^{n+1}_{m+1} : C^{n+1}_{m} = 5 : 5$ $\;$ and $\;$ $C^{n+1}_{m} : C^{n+1}_{m-1} = 5 : 3$
Consider $\;\;\;$ $C^{n+1}_{m+1} : C^{n+1}_{m} = 5 : 5$
i.e. $\;$ $\dfrac{\left(n + 1\right)!}{\left(m + 1\right)! \left(n + 1 - m - 1\right)!} : \dfrac{\left(n + 1\right)!}{m! \left(n + 1 - m\right)!} = 1$
i.e. $\;$ $\dfrac{\left(n + 1\right)!}{\left(m + 1\right)! \left(n - m\right)!} \times \dfrac{m! \left(n + 1 - m\right)!}{\left(n + 1\right)!} = 1$
i.e. $\;$ $\dfrac{m! \left(n + 1 - m\right) \left(n - m\right)!}{\left(m + 1\right) m! \left(n - m\right)!} = 1$
i.e. $\;$ $n + 1 - m = m + 1$
i.e. $\;$ $n = 2m$ $\;\;\; \cdots \; (1)$
Consider $\;\;\;$ $C^{n+1}_{m} : C^{n+1}_{m-1} = 5 : 3$
In view of $(1)$ this becomes
$C^{2m + 1}_{m} : C^{2m + 1}_{m - 1} = 5 : 3$
i.e. $\;$ $\dfrac{\left(2m + 1\right)!}{m! \left(2m + 1 - m\right)!} : \dfrac{\left(2m + 1\right)!}{\left(m - 1\right)! \left(2m + 1 - m + 1\right)!} = \dfrac{5}{3}$
i.e. $\;$ $\dfrac{\left(m - 1\right)! \left(m + 2\right)!}{m! \left(m + 1\right)!} = \dfrac{5}{3}$
i.e. $\;$ $\dfrac{\left(m - 1\right)! \left(m + 2\right) \left(m + 1\right)!}{m \left(m -1\right)! \left(m + 1\right)!} = \dfrac{5}{3}$
i.e. $\;$ $\dfrac{m + 2}{m} = \dfrac{5}{3}$
i.e. $\;$ $3m + 6 = 5m$
i.e. $\;$ $2m = 6$ $\implies$ $m = 3$
Substituting the value of $m$ in $(1)$ gives
$n = 2m = 6$