5. Assign 5 students to three dormitories $A$, $B$, and $C$, with at least 1 student in each dormitory. Among them, student Jia cannot be assigned to dormitory $A$. Then the number of different allocation schemes is ) kinds.
(A) 76
(B) 100
(C) 132
(D) 150
5. B.

First, consider the allocation schemes for three dormitory slots. Since $5=3+1+1=2+2+1$, the problem can be divided into the following categories based on the allocation of slots.
(1) $5=3+1+1$.
$A$ dormitory 3 people, $B$ dormitory 1 person, $C$ dormitory 1 person, there are $\mathrm{C}_{4}^{3} \mathrm{C}_{2}^{1}$ ways;
$A$ dormitory 1 person, $B$ dormitory 3 people, $C$ dormitory 1 person, there are $\mathrm{C}_{4}^{1} \mathrm{C}_{4}^{3}$ ways;
$A$ dormitory 1 person, $B$ dormitory 1 person, $C$ dormitory 3 people, there are $C_{4}^{1} C_{4}^{1}$ ways.
(2) $5=2+2+1$.
$A$ dormitory 1 person, $B$ dormitory 2 people, $C$ dormitory 2 people, there are $\mathrm{C}_{4}^{1} \mathrm{C}_{2}^{2}$ ways;
$A$ dormitory 2 people, $B$ dormitory 1 person, $C$ dormitory 2 people, there are $\mathrm{C}_{4}^{2} \mathrm{C}_{3}^{1}$ ways;
$A$ dormitory 2 people, $B$ dormitory 2 people, $C$ dormitory 1 person, there are $\mathrm{C}_{4}^{2} \mathrm{C}_{3}^{2}$ ways.
Adding all of them together gives 100 ways.