Step 1: Set up the plan.
Not-injective functions $=$ all functions $-$ injective functions. We count both and subtract.
Step 2: Count all functions from $A$ to $B$.
Set $A$ has $3$ elements and $B$ has $4$. Each of the $3$ inputs can map to any of the $4$ outputs, giving $4 \times 4 \times 4 = 4^3 = 64$ functions.
Step 3: Recall what injective needs.
An injective (one-to-one) function sends different inputs to different outputs. So the $3$ outputs must all be distinct.
Step 4: Count the injective functions.
The first input has $4$ choices, the second $3$ (one used up), the third $2$. That is $4 \times 3 \times 2 = 24$.
Step 5: Subtract to find not-injective.
Not-injective $= 64 - 24 = 40$.
Step 6: State the answer.
So $40$ functions from $A$ to $B$ are not injective. \[ \boxed{40} \]