A radioactive nucleus with mass number A and atomic number Z undergoes decay by emitting 3 alpha particles and 2 positrons. Determine the number of protons and neutrons in the final nucleus.

For the initial radioactive nucleus with mass number $A$ and atomic number $Z$:
Number of protons (initial) = $Z$
Number of neutrons (initial) = $A - Z$

An $\alpha$-particle is a helium nucleus ($^4_2\text{He}$).
Each $\alpha$-particle emission reduces the mass number by 4 and the atomic number by 2.
For 3 $\alpha$-particles:
Change in mass number = $3 \times 4 = 12$
Change in atomic number = $3 \times 2 = 6$
After 3 $\alpha$-emissions, the intermediate nucleus has:
New mass number ($A'$) = $A - 12$
New atomic number ($Z'$) = $Z - 6$

A positron ($\beta^+$ particle) is represented as $^0_1\text{e}$.
Each positron emission converts a proton into a neutron, so the mass number remains unchanged, but the atomic number decreases by 1.
For 2 positrons:
Change in mass number = $2 \times 0 = 0$
Change in atomic number = $2 \times 1 = 2$ (atomic number decreases by 2)
After 2 positron emissions, the final nucleus has:
Final mass number ($A''$) = $A' - 0 = A - 12$
Final atomic number ($Z''$) = $Z' - 2 = (Z - 6) - 2 = Z - 8$

Number of protons in the final nucleus = $Z'' = Z - 8$
Number of neutrons in the final nucleus = $A'' - Z'' = (A - 12) - (Z - 8)$

Ratio = $\frac{\text{Number of neutrons}}{\text{Number of protons}}$

Comparing this result with the given options, it matches option (C).

The final answer is $\text{(C)}$.