Answer
a) $6$
b) $10$
c) $0$
d) $2$
e) $1$
Work Step by Step
a) If $n=3$ and $l=1$, there are three possible values for magnetic quantum number: $−1$, $0$ and $1$, and for every of these options there are two possible spin quantum numbers ($+\frac{1}{2}$ and $−\frac{1}{2}$). In total, there are $3×2=6$ electrons which have the required values of principal and angular quantum number.
b) If $n=3$ and $l=2$, allowed values for magnetic quantum number, $m_{l}$, are: $-2$, $-1$, $0$, $1$ and $2$. For each of these options, there are two possible spin quantum numbers ($+\frac{1}{2}$ and $−\frac{1}{2}$). In total, there are $5\times2=10$ electrons which have the required values of principal and angular quantum number.
c) There are $0$ electrons with this combination of quantum numbers, because magnetic quantum number is not in the allowed range, which is between $-l$ and $l$. Since $l=0$, the only option for magnetic quantum number would be $m_{l}=0$.
d) Since principal, angular and magnetic quantum numbers are all in the allowed range of values, there are only $2$ electrons with this combination of quantum numbers (one of these electrons has spin quantum number equal to $+\frac{1}{2}$, and the other one equal to $-\frac{1}{2}$).
e) Since all of the quantum numbers given are correct (in terms of allowed range), there can be only one electron with such combination.
