For a given '$\mathrm{n}$' value $\mathrm{l}$ will have $0$ to $\left(\mathrm{n}-1\right)$ values and for a given '$\mathrm{l}$' value $\mathrm{m}$ will have $-\mathrm{m}$ to $+\mathrm{m}$ values

For $\mathrm{n}=3, \mathrm{l}=3,$ is not possible.Similarly for $\mathrm{n}=2, \mathrm{l}=2$ is also not possible, as $\mathrm{n}$ and $\mathrm{l}$ have same values.

So, set $\mathrm{A}, \mathrm{D}$ are not possible.

If $\mathrm{n}=3,$ $\mathrm{l}$ can have $0$ to $2$ values and '$\mathrm{m}$' can have $-2$ to $+2$ values,

So, set $\mathrm{B}$ is possible.

If $\mathrm{n}=2, \text{'}\mathrm{l}\text{'} \mathrm{can} \mathrm{have} 0 \mathrm{and} 1 \mathrm{values} \mathrm{and} \text{'}\mathrm{m}\text{'} \mathrm{can} \mathrm{have} -1 \mathrm{to} +1 \mathrm{values}.$

So, set C is also possible.

Hence, set $\mathrm{B}$ and $\mathrm{C}$ are the correct set of quantum nmbers.